ECE 45 Homework 4 Solutions

Transcription

1 Spring 26 ECE 45 Homework 4 Solutions Problem. The signal y(t) is generated by convolving a band-limited signal x (t) with another bandlimited signal x 2(t), that is, where, y(t) = x (t) x 2 (t) X (ω) = X 2 (ω) = for ω > π for ω > 2π Impulse train sampling is performed on y(t) to obtain + y p (t) = y(nt) δ(t nt) n= Specify the range of values for the sampling period T which ensures that y(t) is recoverable from y p(t). Using the properties of the Fourier transform, we obtain Y(ω) = X (ω)x 2 (ω) Therefore, Y(ω) = for ω > π This implies that the Nyquist rate for y(t) is 2 x π. Therefore, the sampling period T can at most be 2π 2π = 3 sec. Therefore we have to use T < 3 sec in order to be able to recover y(t) from y p(t). Problem 2. The sampling theorem, as we have derived it, states that a signal x(t) must be sampled at a rate higher than its bandwidth (or equivalently, a rate greater than its highest frequency). This implies that if x(t) has a spectrum as indicated in Figure 2.a then x(t) must be sampled at a rate greater than 2ω 2. However, since the signal has most of its energy concentrated in a narrow band, it would seem reasonable to expect that a sampling rate lower than twice the highest frequency could be used. When x(t) is real, a procedure for bandpass sampling and reconstruction is as outlined in this problem. It consists of multiplying x(t) by a complex-exponential and then sampling the product. The sampling system is shown in Figure 2.b. With x(t) real and with X(ω) nonzero only for ω < ω < ω 2, the frequency is chosen to be ω = (ω + ω 2 ), and the lowpass filter H (ω) has cutoff frequency ( 2 ) (ω 2 ω ). (a) For X(ω) as shown in Figure 2.a, sketch X p (ω). (b) Determine the maximum sampling period T such that x(t) is recoverable from x p(t). (c) Determine a system to recover x(t) from x p(t).

2 (a) Let X (ω) denote the Fourier transform of the signal x (t) obtained by multiplying x(t) with e jω ot. Let X 2 (ω) be the Fourier transform of the signal x 2(t) obtained at the output of the lowpass filter. Then, X (ω), X 2 (ω), and X p (ω) are as shown in Figure S2.a. (b) the Nyquist rate for the signal x2(t) is 2 ω 2 ω = ω 2 ω. Therefore, the sampling period T must be at most 2π in order to avoid aliasing. (ω 2 ω ) 2 (c) A system that can be used to recover x(t) from x p(t) is shown in Figure S2.b.

3 Problem 3. The frequency which, under the sampling theorem, must be exceeded by the sampling frequency is called the Nyquist rate. Determine the Nyquist rate corresponding to each of the following signals: (a) 3 cos 5πt + sin 3πt cos πt (b) sinc(2t) (c) sinc 2 (2t) (d) sinc(2t) + sinc 2 (2t) Note: sinc(x) sin x x (a) The frequencies of the signals are 25 Hz, 5 Hz, and 5 Hz. Since the highest frequency is 5 Hz, the Nyquist rate is 3 Hz. (b) F{sinc(2t)} = f rect ( ) which has a bandwidth of Hz. Hence the Nyquist rate is 2 Hz. 2 2 (c) F{sinc 2 (2t)} = f tri ( ) which has a bandwidth of 2 Hz. Hence the Nyquist rate is 4 Hz. 2 2 Also, note that sinc 2 (2t) is the multiplication of 2 sinc(2t) functions with each other. Hence the Fourier transform is the convolution of the 2 rect functions as found in part (b) above. This implies that the bandwidth is doubled. (Verify using convolution property). (d) Since Fourier transformation is a linear operation, the Fourier transform of sinc(2t) + sinc 2 (2t) is the sum of Fourier transforms of the individual components. As found in parts b and c above, the highest frequency is 2 Hz. Hence the Nyquist rate is 4 Hz.

4 Problem 4. Let x(t) be a signal with Nyquist rate ω o. Determine the Nyquist rate for each of the following signals: (a) x(t) + x(t ) (b) dx(t) dt (c) x 2 (t) (d) x(t) cos(ω o t) If the signal x(t) has Nyquist rate of ω o, then its Fourier transform X(ω) = for ω > ω o/2. Let signal x(t) have a Fourier transform X(ω), i.e., x(t) F X(ω) (a) Using the Fourier transform properties, y(t) = x(t) + x(t ) F Y(ω) Y(ω) = X(ω) + e jω X(ω) = X(ω)( + e jω ) We can only guarantee that Y(ω) = for ω > ω o. Therefore, the Nyquist rate for y(t) is also ωo. 2 (b) Using the Fourier transform properties, y(t) = dx(t) dt F Y(ω) Y(ω) = jωx(ω) We can only guarantee that Y(ω) = for ω > ω o. Therefore, the Nyquist rate for y(t) is also ωo. 2 (c) Using the Fourier transform properties, y(t) = x 2 (t) F Y(ω) Y(ω) = [X(ω) X(ω)] 2π We can only guarantee that Y(ω) = for ω > ω o. Therefore, the Nyquist rate for y(t) is 2ω o. (d) Using the Fourier transform properties, y(t) = x(t) cos(ω o t) F Y(ω) Y(ω) = 2 [X(ω ω o) X(ω + ω o )] We can only guarantee that Y(ω) = for ω > ω o + ω o. Therefore, the Nyquist rate for y(t) is 3ωo. 2

5 Problem 5. Compute the convolution of each of the following pairs of signals x(t) and h(t) by calculating X(ω) and H(ω), using the convolution property, and inverse transforming. (a) x(t) = te 2t u(t), h(t) = e 4t u(t) (b) x(t) = te 2t u(t), h(t) = te 4t u(t) (c) x(t) = e t u(t), h(t) = e t u( t) (a) We have Taking the inverse Fourier transform, we obtain Y(ω) = X(ω)H(ω) = [ (2 + jω) 2] [ (4 + jω) ] = jω jω + 2 (2 + jω) 2 y(t) = 4 e 4t u(t) 4 e 2t u(t) + 2 e 2t u(t) (b) We have Y(ω) = X(ω)H(ω) = [ (2 + jω) 2] [ (4 + jω) 2] = jω + 4 (2 + jω) jω + 4 (4 + jω) 2 Taking the inverse Fourier transform, we obtain y(t) = 4 e 2t u(t) + 4 te 2t u(t) 4 e 4t u(t) + 4 te 4t u(t) (c) We have Taking the inverse Fourier transform, we obtain Y(ω) = X(ω)H(ω) = [ ( + jω) ] [ ( jω) ] = 2 + jω + 2 jω y(t) = 2 e t

6 Problem 6. For x(t) and h(t) as shown in Figure 6, find x(t)*h(t). The convolution is given by x(t) h(t) = x(τ)h(t τ) dτ Hence, inverting h(t) and shifting it [as denoted by h(t τ)], we find the overlap region between x(τ) and h(t τ) and find the integral. t <, t <, t y(t) = 2 e τ dτ = 2(e t e ) t y(t) = 2 e τ dτ + 2 e τ dτ = (4 2e 2e t )

7 t < 2, 2 t < 3, y(t) = 2 e τ dτ + 2 e τ dτ = 4( e ) y(t) = 2 e τ dτ + 2 e τ dτ = 4 2e 2e t 3 t 3 3 t < 4, elsewhere, y(t) = y(t) = 2 e τ dτ = 2(e 3 t e ) t 3 Therefore, finally we have y(t) = {, t < 2(e t e ), t < 4 2e 2e t, t < 4( e ), t < 2 4 2e 2e t 3, 2 t < 3 2(e 3 t e ), 3 t < 4, t > 4

8 Problem 7. For x(t) and h(t) as shown in Figure 7, find x(t)*h(t). The convolution is given by y(t) = x(t) h(t) = x(τ)h(t τ) dτ Hence, inverting h(t) and shifting it [as denoted by h(t τ)], we find the overlap region between x(τ) and h(t τ) and find the integral. t < results in no overlap and hence the integral evaluates to zero i.e. y(t) = If t > and t 2 <, t y(t) = ( 2)e τ dτ = 2(e (t ) )

9 If t 2 > and t 3 <, t 2 t y(t) = (+2)e τ dτ + ( 2)e τ dτ = 2( + e (t ) 2e (t 2) ) t 2 If t 3 >, y(t) = t 2 (+2)e τ dτ + ( 2)e τ dτ = 2(e (t ) 2e (t 2) + e (t 3) ) t 3 Therefore, we finally have y(t) = t t 2, t < 2(e (t ) ), t 2 2( + e (t ) 2e (t 2) ), 2 t 3 { 2(e (t ) 2e (t 2) + e (t 3) ), t 3