Chapter 2.1 Consider the signals displayed in Figure P2.1.Show that each of these signals can be expressed as the sum of rectangular Π (t ) and/or triangular Λ (t ) pulses Figure P2.1 Solution: a ⎛t⎞ ⎛t⎞ x1 (t ) = Π ⎜ ⎟ + Π ⎜ ⎟ ⎝2⎠ ⎝4⎠ ⎛t −3⎞ ⎛t −3⎞ b x2 (t ) = 2Λ ⎜ ⎟− Λ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ c ⎛ 11 ⎞ ⎛ 7⎞ ⎛ 3⎞ ⎛ 1⎞ ⎛ 5⎞ ⎛ 9⎞ x3 (t ) = + Π ⎜ t + ⎟ + Π ⎜ t + ⎟ + Π ⎜ t + ⎟ + Π ⎜ t − ⎟ + Π ⎜ t − ⎟ + Π ⎜ t − ⎟ + 2⎠ ⎝ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ = ∞ ∑ Π ⎡⎣t − ( 2n + 0.5)⎤⎦ n =−∞ © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part ⎛t+4⎞ ⎛t⎞ ⎛t−4⎞ d x4 (t ) = + Λ ⎜ ⎟ + Λ⎜ ⎟ + Λ⎜ ⎟ + ⎝ ⎠ ⎝2⎠ ⎝ ⎠ ∞ ⎛ t − 4n ⎞ = ∑ Λ⎜ ⎟ ⎝ ⎠ n =−∞ 2.2 For the signal x2 (t ) in Figure P2.1 (b) plot the following signals x2 (t − 3) a b x2 (−t ) c x2 (2t ) d x2 (3 − 2t ) Solution: x2 (t − 3) x2 (−t ) 1 0 −6 −5 −4 −3 −2 −1 t x2 (3 − 2t ) x2 (2t ) 1 t −2−1 t 2.3 Plot the following signals a x1 (t ) = 2Π (t / 2) cos(6π t ) ⎡1 ⎤ b x2 (t ) = ⎢ + sgn ( t ) ⎥ ⎣2 ⎦ © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part x3 (t ) = x2 (−t + 2) c d x4 (t ) = sinc ( 2t ) Π ( t / ) Solution: x2 (t ) 1.5 x1(t) 0.5 0 t -0.5 -1 -1.5 0.8 -2 -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 0.6 xx4(t) (t) tt 0.4 x3 (t ) 0.2 -0.2 t -0.4 -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 tt 2.4 Determine whether the following signals are periodic For periodic signals, determine the fundamental period a x1 (t ) = sin(π t ) + 5cos(4π t / 5) Solution: sin(π t ) is periodic with period T1 = 2π π = cos(4π t / 5) is periodic with period T 2π = x1 (t ) is periodic if the ratio can be written as ratio of 4π / T2 integers In the present case, T2 = T1 × = = T2 5 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part Therefore, x1 (t ) is periodic with fundamental period To such that To = 5T1 = 4T2 = 10 b x2 (t ) = e j 3t + e j 9t + cos(12t ) Solution: 2π Similarly, e j 9t is periodic with period 2π 2π π T2 = cos(12t ) is periodic with period T3 = = x2 (t ) is periodic with 12 2π fundamental period To = LCM (T1 , T2 , T3 ) = e j 3t is periodic with period T1 = c x3 (t ) = sin(2π t ) + cos(10t ) Solution: 2π = cos(10t ) is periodic with period 2π T 2π π T2 = = x3 (t ) is periodic if the ratio can be written as ratio of integers 10 T2 In the present case, sin(2π t ) is periodic with period T1 = T1 1× 5 = = T2 π π Since π is an irrational number, the ratio is not rational Therefore, x3 (t ) is not periodic π⎞ ⎛ d x4 (t ) = cos ⎜ 2π t − ⎟ + sin(5π t ) 4⎠ ⎝ Solution: π⎞ 2π ⎛ cos ⎜ 2π t − ⎟ is periodic with period T1 = = sin(5π t ) is periodic with 4⎠ 2π ⎝ T 2π period T2 = = x4 (t ) is periodic if the ratio can be written as ratio of T2 5π integers In the present case, © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part T1 1× 5 = = T2 2 Therefore, x4 (t ) is periodic with fundamental period To such that To = 2T1 = 5T2 = 2.5 Classify the following signals as odd or even or neither a x(t ) = −4t Solution: x(−t ) = 4t = −(−4t ) = − x(t ) So x(t ) is odd b x(t ) = e −t Solution: e c −t =e − −t So x(t ) is even x(t ) = 5cos(3t ) Solution: Since cos(t ) is even, 5cos(3t ) is also even π d x(t ) = sin(3t − ) Solution: π x(t ) = sin(3t − ) = − cos(3t ) which is even e x(t ) = u (t ) Solution: u (t ) is neither even nor odd; For example, u (1) = but u (−1) = ≠ −u (1) © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part f x(t ) = sin(2t ) + cos(2t ) Solution: x(t ) is neither even nor odd; For example, x(π / 8) = but x(−π / 8) = ≠ − x(π / 8) 2.6 Determine whether the following signals are energy or power, or neither and calculate the corresponding energy or power in the signal: a x1 (t ) = u (t ) Solution: The normalized average power of a signal x1 (t ) is defined as T /2 1 u (t ) dt = lim ∫ T →∞ T T →∞ T −T /2 Px1 = lim T /2 1T = T →∞ 2 ∫ dt = lim T Therefore, x1 (t ) is a power signal b x2 (t ) = cos(2π t ) + 3cos ( 4π t ) Solution: cos(2π t ) is a power signal 3cos ( 4π t ) is also a power signal Since x2 (t ) is sum of two power signals, it is a power signal T /2 T /2 2 1 ⎡⎣ cos(2π t ) + 3cos ( 4π t ) ⎤⎦ dt = lim ⎡⎣ cos(2π t ) + 3cos ( 4π t ) ⎤⎦ dt Px2 = lim ∫ ∫ T →∞ T T →∞ T −T /2 −T /2 T /2 ⎡16 cos (2π t ) + cos ( 4π t ) + 24 cos(2π t ) cos ( 4π t ) ⎤⎦dt T →∞ T ∫ ⎣ −T /2 = lim Now T /2 T /2 8T =8 16 cos (2π t )dt = lim + cos(4π t ) ]dt = lim [ ∫ ∫ T →∞ T T →∞ T T →∞ T −T /2 −T /2 lim T /2 lim cos (4π t )dt = 4.5 T →∞ T ∫ −T /2 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part T /2 T /2 24 24 cos(2π t ) cos ( 4π t )dt = lim ∫ ∫ ⎡⎣cos(6π t ) + cos ( 2π t )⎤⎦dt = T →∞ T T →∞ T −T /2 −T /2 lim Therefore, Px2 = + 4.5 = 12.5 c x3 (t ) = t Solution: ⎛ T /2 ⎞ ⎛ −2 ⎞ −2 t dt lim = ⎜− ⎟⎟ = Tlim ⎜ ⎟=0 T →∞ ∫ T →∞ ∫ T →∞ ⎜ →∞ t T / ⎝ ⎠ T /2 − −T /2 −T /2 ⎝ ⎠ T /2 T /2 T /2 ⎞ 1 1⎛ 1 ⎛ −2 ⎞ −2 Px3 = lim t dt t dt 1/ lim lim = = − = lim ⎜ ⎜ ⎟ ⎟=0 ⎟ →∞ T →∞ T ∫ T →∞ T ∫ T →∞ T ⎜ T T ⎝T / 2⎠ −T /2 −T /2 ⎝ t −T /2 ⎠ T /2 Ex3 = lim T /2 1/ t dt = lim x3 (t ) is neither an energy nor a power signal d x4 (t ) = e−α t u (t ) Solution: T /2 Ex4 = lim T →∞ ∫ T /2 −α t e u (t ) dt = lim T →∞ −T /2 ∫ e −2α t ⎛ e −2α t dt = lim ⎜ − T →∞ ⎜ 2α ⎝ T /2 ⎞ 1 ⎟= lim ( −e −α T + 1) = ⎟ 2α T →∞ 2α ⎠ Thus x4 (t ) is an energy signal e x5 (t ) = Π (t / 3) + Π (t ) Solution: x5 (t ) −3 / 3/ t © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part T /2 Ex5 = lim T →∞ ∫ Π (t / 3) + Π (t ) dt = −T /2 −1/2 1/ −3/ −1/2 ∫ 1dt + ∫ −1/ 4dt + ∫ 1dt −3/2 = 1+ +1 = Thus x5 (t ) is an energy signal f x6 (t ) = 5e( −2t + j10π t )u (t ) Solution: T /2 Ex6 = lim T →∞ ∫ 5e( −2t + j10π t )u (t ) dt = lim T →∞ −T /2 ⎛ e = 25 lim ⎜ − T →∞ ⎜ ⎝ −4 t T /2 T /2 ∫ T /2 5e −2t e j10π t ) dt = lim 25 ∫ e−4t dt T →∞ ⎞ 25 25 ⎟= lim ( −e −2T + 1) = →∞ T ⎟ 4 ⎠ Thus x6 (t ) is an energy signal g x7 (t ) = ∞ ∑ Λ ⎡⎣( t − 4n ) / 2⎤⎦ n =−∞ Solution: x7 (t ) is a periodic signal with period To = Px7 = To To / 2 ⎡⎣ Λ ( t / ) ⎤⎦ dt = ∫ To −To / 2 1 ( ) ∫0 (1 − t ) dt = ∫0 − 2t + t dt t3 ⎞ 1⎛ 1⎛ 1⎞ = ⎜ t − t + ⎟ = ⎜1 − + ⎟ = 2⎝ ⎠0 2⎝ 3⎠ 2.7 Evaluate the following expressions by using the properties of the delta function: a x1 (t ) = δ (4t ) sin(2t ) Solution: δ (4t ) = δ (t ) 1 x1 (t ) = δ (t ) sin(2t ) = δ (t ) sin(0) = 4 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part π⎞ ⎛ b x2 (t ) = δ (t ) cos ⎜ 30π t + ⎟ 4⎠ ⎝ Solution: π⎞ π⎞ ⎛ ⎛ ⎛π ⎞ x2 (t ) = δ (t ) cos ⎜ 30π t + ⎟ = δ (t ) cos ⎜ + ⎟ = cos ⎜ ⎟ δ (t ) 4⎠ 4⎠ ⎝ ⎝ ⎝4⎠ c x3 (t ) = δ (t )sinc(t + 1) Solution: x3 (t ) = δ (t )sinc(t + 1) = δ (t )sinc(0 + 1) = δ (t )sinc(1) = δ (t ) × = d x4 (t ) = δ (t − 2)e − t sin(2.5π t ) Solution: x4 (t ) = δ (t − 2)e − t sin(2.5π t ) = δ (t − 2)e −2sin(2.5π × 2) = δ (t − 2)e −2sin(π ) = ∞ e ∫ δ (2t )sinc(t )dt x5 (t ) = −∞ Solution: ∞ ∞ x5 (t ) = ∫ δ (2t )sinc(t )dt = ∫ δ (t )sinc(t ) dt −∞ −∞ ∞ ∞ 1 δ (t )sinc(0)dt = ∫ δ (t )dt = ∫ −∞ −∞ = ∞ f x6 (t ) = ∫ δ (t − 3) cos(t )dt −∞ Solution: x6 (t ) = ∞ ∞ −∞ −∞ ∫ δ (t − 3) cos(t )dt = cos(3) ∫ δ (t − 3)dt = cos(3) © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part ∞ ∫ δ (2 − t ) − t g x7 (t ) = dt −∞ Solution: ∞ x7 (t ) = ∫ δ (2 − t ) −∞ ∞ 1 dt = ∫ δ (t − 2) dt 1− t 1− t3 −∞ ∞ ∞ 1 dt = − ∫ δ (t − 2)dt = − = ∫ δ (t − 2) 1− −∞ −∞ ∞ h x8 (t ) = ∫ δ (3t − 4)e −3t dt −∞ Solution: ∞ x8 (t ) = −3t ∫ δ (3t − 4)e dt = −∞ ∞ ∞ ⎛ 4⎞ δ ⎜ t − ⎟ e −3t dt ∫ −∞ ⎝ ⎠ e −4 ⎛ ⎞ −3× = ∫ δ ⎜ t − ⎟ e dt = −∞ ⎝ ⎠ i ∞ ⎛ e −4 4⎞ ∫ δ ⎜ t − ⎟⎠ dt = −∞ ⎝ x9 (t ) = δ '(t ) ⊗ Π (t ) Solution: ∞ x9 (t ) = d ∫ Π (t − τ )δ '(τ )dτ = (−1) dτ Π (t − τ ) τ −∞ =0 = Π (t ) = δ ( t + 0.5 ) − δ ( t − 0.5 ) ' 2.8 For each of the following continuous-time systems, determine whether or not the system is (1) linear, (2) time-invariant, (3) memoryless, and (4) casual a y (t ) = x(t − 1) Solution: The system is linear, time-invariant, causal, and has memory The system has memory because current value of the output depends on the previous value of the input The system is causal because current value of the output does not depend on future inputs To prove linearity, let x(t ) = α x1 (t ) + β x2 (t ) The response of the system to x(t ) is 10 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part b v(t ) = x(t ) cos(100π t ) Solution: ⎡⎣ x(t )e j100π t + x(t )e− j100π t ⎤⎦ v(t ) = x(t ) cos(100π t ) = Now ⎡⎣ x(t )e j100π t + x(t )e− j100π t ⎤⎦ ℑ ←⎯→V ( f ) = ⎡⎣ X ( f − 50 ) + X ( f + 50 ) ⎤⎦ 2 1 = + j 2π ( f − 50 ) + j 2π ( f + 50 ) + After simplification, we get V( f ) = j 2π f + j 20π f + (π 2104 + 25 − 4π f ) c v(t ) = x(t )e j10t Solution: 5⎞ ⎛ ℑ →V ( f ) = X ⎜ f − ⎟ = v(t ) = x(t )e j10t ←⎯ 5⎞ π⎠ ⎛ ⎝ j 2π ⎜ f − ⎟ + π⎠ ⎝ dx(t ) d v(t ) = dt Solution: Using the differentiation property of FT V ( f ) = j 2π fX ( f ) = d ℑ x(t ) ←⎯→ j 2πfX ( f ) , we obtain dt j 2π f + j 2π f e v(t ) = x(t ) ⊗ u (t ) Solution: ℑ Using the convolution property of FT x(t ) ⊗ y (t ) ←⎯→ X ( f )Y ( f ) , we can write 33 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part ⎛1 1 ⎞ ⎜ δ( f )+ ⎟ j 2π f ⎠ + j 2π f ⎝ 1 = δ(f ) + + j 2π f j 2π f ( + j 2π f ) V ( f ) = X ( f )U ( f ) = = 1 δ(f )+ 10 j 2π f ( + j 2π f ) 2.21 Consider the delay element y (t ) = x(t − 3) a What is the impulse response h(t)? Solution: Since y (t ) = h(t ) ⊗ x(t ) = x(t − 3) , the impulse response of the delay element is given from (2.16) as h(t ) = δ (t − 3) b What is the magnitude and phase response function of the system? Solution: H ( f ) = ℑ{δ (t − 3)} = e − j 6π f H( f ) =1 H ( f ) = −6π f 2.22 The periodic input x(t ) to an LTI system is displayed in Figure P2.7 The frequency response function of the system is given by H( f ) = 2 + j 2π f Figure P2.7 34 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part a Write the complex exponential FS of input x(t ) Solution: The input x(t ) is rectangular pulse train in Example 2.24 shifted by To / and duty cycle τ = 0.5 That is, To x(t ) = gTo ( t − To / ) = ∞ ⎡ ( t − nTo − To / ) ⎤ ⎥ 0.5To ⎣ ⎦ ∑ Π⎢ n =−∞ The complex exponential FS of gTo (t ) from Example 2.24 with To = ( f o = 0.5 ) and τ To = 0.5 is given by ∞ gTo (t ) = 0.5 ∑ sinc ( 0.5n )e jπ nt n =−∞ In Exercise 2.15(a), we showed that time shifting introduces a linear phase shift in the FS coefficients; their magnitudes are not changed The phase shift is equal to e − j 2π nfou for a time shift of u The exponential FS coefficients x(t ) are Cn = 0.5sinc ( 0.5n ) e − j 2π nf o ( 0.25To ) = e − jπ n /2 0.5sinc ( 0.5n ) ∞ x(t ) = 0.5 ∑ ⎡⎣e − jπ n /2sinc ( 0.5n ) ⎤⎦e jπ nt n =−∞ b Plot the magnitude and phase response functions for H ( f ) Solution: H( f ) = = + j 2π f + jπ f H ( nf o ) = 1 = + jπ f + jπ nf o 35 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 0 -10 -5 Phase Response(degrees) Magnitude Response(dB) -20 -10 -15 -20 -30 -40 -50 -60 -70 -25 -80 -30 Frequency (f) -90 Frequency (f) c Compute the complex exponential FS of the output y (t ) Solution: Using (2.114), the output of an LTI system to the input x(t ) = ∞ ∑ ⎡⎣0.5e − jπ n / n =−∞ sinc ( 0.5n ) ⎤⎦e jπ nt Cn is given by y (t ) = ∞ ∑ C H (0.5n)e π j nt n =−∞ n ⎡ ⎤ ⎢ 0.5e − jπ n /2 ⎥ sinc ( 0.5n ) ⎥ e jπ nt = ∑⎢ n =−∞ ⎢1 + j 05π n ⎥ FS coefficient of y ( t ) ⎣⎢ ⎦⎥ ∞ 2.23 The frequency response of an ideal LP filter is given by ⎧⎪5e − j 0.0025π f , H( f ) = ⎨ ⎪⎩0, f < 1000 Hz f > 1000 Hz Determine the output signal in each of the following cases: a x(t ) = 5sin ( 400π t ) + cos (1200π t − π / ) − cos ( 2200π t + π / ) 36 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part Solution: ⎧⎪5, H( f ) = ⎨ ⎪⎩0, ⎧− ⎪ 0.0025π f , H( f ) = ⎨ ⎪⎩0, Since H (200) = and f < 1000 Hz f > 1000 Hz f < 1000 Hz f > 1000 Hz H (200) = −π / , the output of the system for an input 5sin(400πt) can now be expressed using (2.117) as 25sin(400πt − π/2) Next H (600) = and H (600) = −3π / = π / , the output of the system for an input cos (1200π t − π / ) can now be expressed using (2.117) as 10 cos (1200π t − π / + π / ) = 10 cos (1200π t ) Now H (1100) = So the LP filter doesn’t pass cos ( 2200π t + π / ) The output of the LP filter is therefore given by y (t ) = 25sin ( 400π t − π / ) + 10 cos (1200π t ) b x(t ) = 2sin(400π t ) + sin ( 2200π t ) πt Solution: Since H (200) = and H (200) = −π / , the output of the system for an input 2sin(400πt) is 10sin(400πt − π/2) To calculate the response to sin ( 2200π t ) , we note that πt sin ( 2200π t ) ⎛ f ⎞ ℑ = 2200 × sinc ( 2200t ) ←⎯ →Π⎜ ⎟ πt ⎝ 2200 ⎠ In frequency domain, the output of LP filter to sin ( 2200π t ) is πt ⎛ f ⎞ − j 0.0025π f ⎛ f ⎞ ⎛ f ⎞ − j 0.0025π f Π⎜ Π⎜ ⎟ 5e ⎟ = 5Π ⎜ ⎟e ⎝ 2200 ⎠ ⎝ 2000 ⎠ ⎝ 2000 ⎠ 37 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part Now ⎛ f ⎞ − j 0.0025π f ℑ Π⎜ ←⎯→ 2000sinc ⎡⎣ 2000 ( t − 0.00125 ) ⎤⎦ ⎟e ⎝ 2000 ⎠ The output of the LP filter is therefore given by y (t ) = 10sin ( 400π t − π / ) + 10000sinc ⎡⎣ 2000 ( t − 0.00125 ) ⎤⎦ c x(t ) = cos(400π t ) + sin (1000π t ) πt Solution: Since H (200) = and H (200) = −π / , the output of the system for an input cos(400π t ) is 5cos ( 400π t − π / ) To calculate the response to sin (1000π t ) , we note that πt sin (1000π t ) ⎛ f ⎞ ℑ = 1000 × sinc (1000t ) ←⎯ →Π ⎜ ⎟ πt ⎝ 1000 ⎠ In frequency domain, the output of LP filter to sin (1000π t ) is πt ⎛ f ⎞ ⎛ f ⎞ − j 0.0025π f 5e− j 0.0025π f Π ⎜ ⎟ = 5Π ⎜ ⎟e ⎝ 1000 ⎠ ⎝ 1000 ⎠ Now ⎛ f ⎞ − j 0.0025π f ℑ Π⎜ ←⎯→1000sinc ⎡⎣1000 ( t − 0.00125 ) ⎤⎦ ⎟e ⎝ 1000 ⎠ The output of the LP filter is therefore given by y (t ) = 5cos ( 400π t − π / ) + 5000sinc ⎡⎣1000 ( t − 0.00125 ) ⎤⎦ d x(t ) = 5cos(800π t ) + 2δ (t ) Solution: 38 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part Since H (400) = and H (400) = −π , the output of the system for an input 5cos ( 800π t ) is 25cos ( 800π t − π ) The response of the system to δ (t ) is h(t ) The impulse response of the ideal LP filter is 10000sinc ⎡⎣ 2000 ( t − 0.00125 ) ⎤⎦ Combining y (t ) = 25cos ( 800π t − π ) + 2h(t ) = 25cos ( 800π t − π ) + 20 × 103 sinc ⎡⎣ 2000 ( t − 0.00125 ) ⎤⎦ 2.24 The frequency response of an ideal HP filter is given by ⎧⎪4, H( f ) = ⎨ ⎪⎩0, f > 20 Hz, f < 20 Hz Determine the output signal y (t ) for the input a x(t ) = + cos ( 50π t − π / ) − cos ( 75π t + π / ) Solution: y (t ) = 8cos ( 50π t − π / ) − cos ( 75π t + π / ) b x(t ) = cos ( 20π t − 3π / ) + 3cos (100π t + π / ) Solution: y (t ) = 12 cos (100π t + π / ) 2.25 The frequency response of an ideal BP filter is given by ⎧⎪2e− j 0.0005π f , 900< f < 1000 Hz, H( f ) = ⎨ otherwise ⎪⎩0, Determine the output signal y (t ) for the input a x(t ) = cos (1850π t − π / ) − cos (1900π t + π / ) Solution: H (925) = and H (950) = 39 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part H ( f ) = −0.0005π f Therefore, H (925) = −0.0005π f = −0.0005π × 925 = −0.462π H (950) = −0.0005π f = −0.0005π × 950 = −0.475π y (t ) = cos (1850π t − π / − 0.462π ) − cos (1900π t + π / − 0.475π ) b x(t ) = sinc ( 60t ) cos(1900π t ) Solution: X(f ) = = ⎛ f ⎞ Π ⎜ ⎟ ⊗ [δ ( f − 950) + δ ( f + 950) ] 60 ⎝ 60 ⎠ ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ Π⎜ ⎟+Π⎜ ⎟⎥ ⎢ 120 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦ Y( f )= H( f )X ( f ) = 2e − j 0.0005π f × ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ × ⎢Π ⎜ ⎟+Π⎜ ⎟⎥ 120 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦ Now ⎛ f − 950 ⎞ ℑ j1900π t ×Π⎜ ⎟ ←⎯→ sinc ( 60t ) e 60 ⎝ 60 ⎠ e − j 0.0005π f ⎛ f − 950 ⎞ ℑ j1900π ( t − 0.00025 ) ×Π⎜ ⎟ ←⎯→ sinc ⎡⎣60 ( t − 0.00025 ) ⎤⎦ e 60 ⎝ 60 ⎠ Similarly ⎛ f + 950 ⎞ ℑ − j1900π t ×Π⎜ ⎟ ←⎯→ sinc ( 60t ) e 60 ⎝ 60 ⎠ e− j 0.0005π f ⎛ f + 950 ⎞ ℑ − j1900π ( t − 0.00025 ) ×Π⎜ ⎟ ←⎯→ sinc ⎡⎣ 60 ( t − 0.00025 ) ⎤⎦ e 60 ⎝ 60 ⎠ Therefore, y (t ) = sinc ⎡⎣60 ( t − 0.00025 ) ⎤⎦ ⎡⎣e j1900π ( t −0.00025) + e − j1900π ( t −0.00025) ⎤⎦ = 2sinc ⎡⎣60 ( t − 0.00025 ) ⎤⎦ cos ⎡⎣1900π ( t − 0.00025 ) ⎤⎦ 40 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part c x(t ) = sinc ( 30t ) cos(1900π t ) Solution: X(f ) = = ⎛ f ⎞ Λ ⎜ ⎟ ⊗ [δ ( f − 950) + δ ( f + 950) ] 30 ⎝ 60 ⎠ ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ Λ⎜ ⎟ + Λ⎜ ⎟⎥ ⎢ 60 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦ Y( f )= H( f )X ( f ) = 2e − j 0.0005π f × ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ Λ⎜ ⎟+ Λ⎜ ⎟⎥ ⎢ 60 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦ Now ⎛ f − 950 ⎞ ℑ j1900π t ×Λ⎜ ⎟ ←⎯→ sinc ( 30t ) e 30 ⎝ 60 ⎠ e − j 0.0005π f ⎛ f − 950 ⎞ ℑ j1900π ( t − 0.00025) ×Λ⎜ ⎟ ←⎯→ sinc ⎡⎣30 ( t − 0.00025 ) ⎤⎦ e 30 ⎝ 60 ⎠ Similarly ⎛ f + 950 ⎞ ℑ − j1900π t ×Π⎜ ⎟ ←⎯→ sinc ( 30t ) e 30 ⎝ 60 ⎠ e − j 0.0005π f ⎛ f + 950 ⎞ ℑ − j1900π ( t − 0.00025 ) ×Π⎜ ⎟ ←⎯→ sinc ⎡⎣30 ( t − 0.00025 ) ⎤⎦ e 30 ⎝ 60 ⎠ Therefore, y (t ) = sinc ⎡⎣30 ( t − 0.00025 ) ⎤⎦ ⎡⎣ e j1900π ( t −0.00025) + e− j1900π ( t −0.00025) ⎤⎦ = 2sinc ⎡⎣30 ( t − 0.00025 ) ⎤⎦ cos ⎡⎣1900π ( t − 0.00025 ) ⎤⎦ 2.26 The signal 2e −2t u (t ) is input to an ideal LP filter with passband edge frequency equal to Hz Find the energy density spectrum of the output of the filter Calculate the energy of the input signal and the output signal Solution: 41 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part ∞ Ex = ∫ −∞ ∞ e −4t 2e u (t ) dt = ∫ 4e dt = −4 −2 t ℑ x(t ) = 2e −2t u (t ) ←⎯ → ∞ −4 t =1 2 + j 2π f The energy density spectrum of the output y(t), Y ( f ) , is related to the energy density spectrum of the input x(t) 2 ⎛ f ⎞ Y( f ) = H( f ) X( f ) = Π⎜ ⎟ ⎝ 10 ⎠ + j 2π f 2 ∞ Ey = ∫ ∞ y (t ) dt = −∞ ∫ Y( f ) ∞ 2 ∫ df = −∞ 2 −∞ 2 ⎛ f ⎞ df Π⎜ ⎟ ⎝ 10 ⎠ + j 2π f 1 df = 2∫ df =∫ + jπ f 1+ π f −5 Making change of variables π f = u ⇒ df = Ey = π 5π ∫ 1+ u du = π tan −1 ( u ) 5π = π π du , we obtain (1.507 ) = 0.9594 Thus output of the LP filter contains 96% of the input signal energy 2.27 Calculate and sketch the power spectral density of the following signals Calculate the normalized average power of the signal in each case a x(t ) = cos (1000π t − π / ) − cos (1850π t + π / ) x1 ( t ) x2 ( t ) Solution: T /2 Rx (τ ) = lim x(t ) x(t − τ )dt = Rx1 (τ ) + Rx2 (τ ) − Rx1x2 (τ ) − Rx2 x1 (τ ) T →∞ T ∫ −T /2 Now 42 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part T /2 Rx1 (τ ) = lim x1 (t ) x1 (t − τ )dt T →∞ T ∫ −T /2 π⎞ π⎤ ⎛ ⎡ cos ⎜1000π t − ⎟ cos ⎢1000π ( t − τ ) − ⎥ dt T →∞ T ∫ 2⎠ 2⎦ ⎝ ⎣ −T /2 T /2 = lim T /2 { } cos (1000πτ ) + cos ⎡⎣1000π ( 2t − τ ) − π ⎤⎦ dt T →∞ T ∫ −T /2 = lim = cos (1000πτ ) The second term is zero because it integrates a sinusoidal function over a period Similarly, T /2 1 x2 (t ) x2 (t − τ )dt = cos (1850πτ ) ∫ T →∞ T −T / Rx2 (τ ) = lim The cross-correlation term T /2 T /2 1 x1 (t ) x2 (t − τ ) = lim cos (1000π t − π / ) cos ⎡⎣1850π ( t − τ ) + π / ⎤⎦dt ∫ T →∞ T T →∞ T ∫ −T / −T / Rx1 x2 (τ ) = lim T /2 = lim {cos (850π t − 1850πτ + 3π / ) + sin ( 2850π t − 1850πτ − π / )}dt T →∞ T ∫ −T / is zero because it integrates a sinusoidal function over a period in each case Similarly, it can be shown that all other cross-correlation terms are zero Therefore, Rx (τ ) = Rx1 (τ ) + Rx2 (τ ) = cos (1000πτ ) + cos (1850πτ ) ⎧ ⎫ Gx ( f ) = ℑ{Rx (τ )} = ℑ ⎨2 cos (1000πτ ) + cos (1850πτ ) ⎬ ⎩ ⎭ = [δ ( f − 500) + δ ( f + 500) ] + [δ ( f − 925) + δ ( f + 925) ] The normalized average power is obtained by using (2.172) as 43 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part ∞ ∞ −∞ −∞ ∞ ∫ [δ ( f − 500) + δ ( f + 500)] df + ∫ [δ ( f − 925) + δ ( f + 925)] df ∫ Gx ( f )df = Px = −∞ 1 = 1+1+ + = 4 Power Spectral Density 0.8 0.6 0.4 0.2 -1 -0.8 -0.6 -0.4 -0.2 0.2 Frequency (kHz) 0.4 0.6 0.8 b x(t ) = [1 + sin(200π t ) ] cos(2000π t ) Solution: x(t ) = [1 + sin(200π t ) ] cos(2000π t ) = cos(2000π t ) + sin(200π t ) cos(2000π t ) 1 = cos(2000π t ) + sin ( 2200π t ) − sin (1800π t ) 2 x (t ) x2 ( t ) x3 ( t ) Now T /2 Rx (τ ) = lim x(t ) x(t − τ )dt = Rx1 (τ ) + Rx2 (τ ) + Rx3 (τ ) T →∞ T ∫ −T /2 where cos ( 2000πτ ) Rx2 (τ ) = cos ( 2200πτ ) Rx3 (τ ) = cos (1800πτ ) Rx1 (τ ) = 44 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part Therefore, 1 Rx (τ ) = cos ( 2000πτ ) + cos ( 2200πτ ) + cos (1800πτ ) 8 1 ⎧1 ⎫ Gx ( f ) = ℑ{Rx (τ )} = ℑ ⎨ cos ( 2000πτ ) + cos ( 2200πτ ) + cos (1800πτ ) ⎬ 8 ⎩2 ⎭ 1 = [δ ( f − 1000) + δ ( f + 1000) ] + [δ ( f − 1100) + δ ( f + 1100) ] 16 + [δ ( f − 900) + δ ( f + 900) ] 16 The normalized average power is obtained by using (2.172) as ∞ ∞ ∞ 1 Px = ∫ Gx ( f )df = ∫ [δ ( f − 1000) + δ ( f + 1000) ] df + ∫ [δ ( f − 1100) + δ ( f + 1100)] df −∞ 16 −∞ −∞ ∞ + 1 1 [δ ( f − 900) + δ ( f + 900)] df = + + = ∫ 16 −∞ 8 Power Spectral Density 0.25 0.2 0.15 0.1 0.05 c -1 -0.8 -0.6 -0.4 -0.2 0.2 Frequency (kHz) 0.4 0.6 0.8 x(t ) = cos ( 200π t ) sin (1800π t ) Solution: 45 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 1 ⎡⎣1 + cos ( 400π t ) ⎤⎦ sin (1800π t ) = sin (1800π t ) + sin (1800π t ) cos ( 400π t ) 2 1 = sin (1800π t ) + sin (1400π t ) + sin ( 2200π t ) 4 x(t ) = Now Rx (τ ) = Rx1 (τ ) + Rx2 (τ ) + Rx3 (τ ) where Rx1 (τ ) = cos (1800πτ ) Rx2 (τ ) = cos (1400πτ ) 32 Rx3 (τ ) = cos ( 2200πτ ) 32 Therefore, 1 Rx (τ ) = cos (1800πτ ) + cos (1400πτ ) + cos ( 2200πτ ) 32 32 1 ⎧1 ⎫ Gx ( f ) = ℑ{Rx (τ )} = ℑ ⎨ cos (1800πτ ) + cos (1400πτ ) + cos ( 2200πτ ) ⎬ 32 32 ⎩8 ⎭ 1 = [δ ( f − 900) + δ ( f + 900) ] + [δ ( f − 700) + δ ( f + 700) ] 16 64 + [δ ( f − 1100) + δ ( f + 1100) ] 64 The normalized average power is obtained by using (2.172) as ∞ ∞ ∞ 1 Px = ∫ Gx ( f )df = [δ ( f − 900) + δ ( f + 900)]df + ∫ [δ ( f − 700) + δ ( f + 700)] df ∫ 16 −∞ 64 −∞ −∞ ∞ + 1 1 [δ ( f − 1100) + δ ( f + 1100)] df = + + = ∫ 64 −∞ 32 32 16 46 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part 0.1 0.09 Power Spectral Density 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 47 © 2012 by McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part ... because current value of the output depends on the previous value of the input The system is causal because current value of the output does not depend on future inputs To prove linearity, let x(t... = 3x(t ) − Solution: The system is nonlinear, time-invariant, causal, and memoryless The system is memoryless because current value of the output depends only on the current value of the input... (t ) = x(t ) Solution: The system is nonlinear, time-invariant, causal, and memoryless The system is memoryless because current value of the output depends only on the current value of the input