Homework 4 May 2017 1. An LTI system has an input, x(t) and output y(t) related through the equation y(t) = t e (t t ) x(t 2)dt Determine the impulse response of the system. Rewriting as y(t) = t e (t t ) x(t 2)dt y(t) = Let t 2 = τ in the original equation and y(t) = t 2 h(t τ)x(τ)dτ e (t 2 τ) x(τ)dτ h(t) = e (t 2) u(t 2) 2. Determine the minimum sampling frequency, f s, required to sample the following signals without aliasing (a) x(t) = (sin(1000πt)/πt) y(t) where y(t) is any signal in time domain Let k(t) = sin(1000πt)/πt) { 1 ω < 1000π K(jω) = 0 otherwise Now, X(jω) = K(jω)Y (jω) ω max in X(jω) is 1000π f s 2ω max = 1kHz 1
(b) f(t) = cos(6000πt) + 2 F (jω) = π[δ(ω 6000π) + δ(ω + 6000π)] + 4πδ(ω) (c) h(t) = f(t) sin(1000πt)/πt = ω max = 6000π = f s 2ω max G(jω) = { 1 ω < 1000π 0 ω > 1000π = 6kHz H(jω) = 1 F (jω) G(jω) = 1 2 [G(j(ω 6000π))+G(j(ω+6000π))]+G(jω)) = ω max = 7000π = f s 2ω max = 7kHz (d) x(t) = 3 cos(50πt) + 10 sin(300πt) cos(100πt) The frequencies of the signals are 25Hz, 150Hz and 50Hz. As the highest frequency is 150Hz, the Nyquist rate is 300Hz (e) x(t) = sinc(200πt) Fourier transform of x(t) will be a rectangular pulse with the highest angular frequency, ω max = 200π. Therefore, the Nyquist rate is 400π rad/sec or 200Hz. Note: Use the FT of K(jω) from part a) with appropriate scaling. (f) x(t) = sinc 2 (200πt) Let k(t) = sinc(200πt), then X(jω) = K(jω) K(jω) K(jω) is a rectangular pulse of bandwidth 400π rad/sec around 0, i.e. ω max = 200π rad/sec for rectangular pulse. The convolution of two rectangular pulse will be a triangular pulse with ω max = 400π rad/sec. Thus, the minimum sampling frequency is 400 Hz. (g) x(t) = sinc(200πt) + sinc 2 (200πt) ω max = 400π rad/sec from part e) and f). Thus minimum sampling frequency is 400 Hz. 3. The signal y(t) is generated by convolving a band-limited signal x 1 (t) with another band limited signal x 2 (t), that is, y(t) = x 1 (t) x 2 (t) 2
where, X 1 (ω) = 0 for ω > 1000π X 2 (ω) = 0 for ω > 2000π Impluse train sampling is performed on y(t) to obtain y p (t) = n= y(nt )δ(t nt ) Specify the range of values for the sampling period T which ensures that y(t) is recoverable from y p (t) Using the properties fo the Fourier transform we obtain Therefore Y (ω) = X 1 (ω)x 2 (ω) Y (ω) = 0 for ω > 1000π This implies that the Nyquist rate for y(t) is 2000π rad/sec. Therefore, the sampling period T can at most be 2000π = 10 3 sec. Therefore, T 10 3 sec in order to be able to recover y(t) from y p (t) 4. 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 by X(ω) (in the figure below) then x(t) must be sampled at a rate higher then 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 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 block diagram below. With x(t) real and with X(ω) nonzero only for ω 1 < ω < ω 2, the frequency is chosen to be ω 0 = (ω 1 + ω 2 )/2, and the lowpass filter H 1 (ω) has cutoff frequency ( 1 2 )(ω 2 ω 1 ) 3
(a) For X(ω) as shown in the figure sketch X p (ω) Let X 1 (ω) denote the Fourier transform of the signal x 1 (t) obtained by multiplying x(t) with e jω0t. Let X 2 (ω) be the Fourier transform of the signal x 2 (t) obtained at the output of the lowpass filter. Then, X 1 (ω), X 2 (ω) and X p (ω) are as shown in figure below. (b) Determine the maximum sampling period T such that x(t) is recoverable from x p (t) The Nyquist rate for the signal x 2 (t) is 2 ω2 ω1 2 = ω 2 ω 1. Therefore, the sampling period T must be at most avoid aliasing. ω 2 ω 1 in order to (c) Determine a system to recover x(t) from x p (t) A system that can be used to recover x(t) from x p (t) is shown in figure below 4
Let x 1 (t) be the input to the block Re{.}. X 1 (ω) is twice the left hand side triangular pulse of X(ω). Now, Re(x 1 (t)) = (x 1 (t) + x 1(t))/2. The equivalent operation is frequency domain is (X 1 (ω)+x 1 ( ω))/2. Note that for the given X 1 (ω), (X 1 (ω) + X 1 ( ω))/2 = X 1 (ω)/2 + X 1 ( ω)/2 = X(ω).Thus, x(t) is recovered at the output. 5. Let x(t) be a signal with Nyquist rate ω 0. Determine the Nyquist rate for each of the following signals: (a) x(t) + x(t 1) If the signal x(t) has Nyquist rate of ω 0, then its Fourier transform X(ω) = 0 for ω > ω 0 /2 Let signal x(t) have a Fourier transform X(ω), i.e. x(t) F X(ω)) Using the Fourier transform properties, y(t) = x(t) + x(t 1) F Y (ω) Y (ω) = X(ω) + e jω X(ω) = X(ω)(1 + e jω ) We can only guarantee that Y (ω) = 0 for ω > ω0 2. Therefore, the Nyquist rate for y(t) is also ω 0 (b) dx(t) dt Using the Fourier transform properties, y(t) = d(x(t) dt F Y (ω) Y (ω) = jωx(ω) We can only guarantee that Y (ω) = 0 for ω > ω0 2. Therefore, the Nyquist rate for y(t) is also ω 0. (c) x 2 (t) Using the Fourier transform properties, y(t) = x 2 F Y (ω) Y (ω) = 1 [X(ω) X(ω)] We can only guarantee that Y (ω) = 0 for ω > ω 0. Therefore, the Nyquist rate for y(t) is 2ω 0. 5
(d) x(t)cos(ω 0 t) Using the Fourier transform properties, y(t) = x(t)cos(ω 0 t) F Y (ω) Y (ω) = 1 2 [X(ω ω 0) + X(ω + ω 0 )] We can only guarantee that Y (ω) = 0 for ω > ω 0 + ω0 2. Therefore, the Nyquist rate for y(t) is 3ω 0. 6. Determine the convolution of each of the following pairs of signal x(t) and h(t) by calculating X(ω) and H(ω), and inverse transforming. Now verify your answers by computing the convolution in time domain. (a) x(t) = te 2t u(t), h(t) = e 4t u(t) We have Y (ω) = X(ω)H(ω) = 1 1 (2 + jω) 2 (4 + jω) Y (ω) = 1/4 4 + jω 1/4 2 + jω + 1/2 (2 + jω) 2 Taking the inverse Fourier transform we obtain y(t) = 1 4 e 4t u(t) 1 4 e 2t u(t) + 1 2 te 2t u(t) (b) x(t) = te 2t u(t), h(t) = te 4t u(t) We have 1 Y (ω) = X(ω)H(ω) = [ (2 + jω) 2 ][ 1 (4 + jω) 2 ] Y (ω) = 1/4 4 + jω + 1/4 (4 + jω) 2 + 1/4 2 + jω + 1/4 (2 + jω) 2 Taking the inverse Fourier transform we obtain y(t) = 1 4 te 4t u(t) 1 4 e 4t u(t) + 1 4 e 2t u(t) + 1 4 te 2t u(t) (c) x(t) = e t u(t), h(t) = e t u( t) We have 1 1 Y (ω) = X(ω)H(ω) = (1 + jω) (1 jω) Y (ω) = 0.5 (1 + jω) + 0.5 (1 jω) Taking the inverse Fourier transform we obtain y(t) = 1 2 e t 6
7. x(t) and h(t) are shown in the figure below. 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 1 t < 0 0 t < 1 t y(t) = 2 e τ dτ = 2(e t e 1 ) 1 1 t < 2 0 τ y(t) = 2 e τ dτ + 2 e τ dτ = (4 2e 1 2e t ) 1 0 7
2 t < 3 0 y(t) = 2 e τ dτ + 2 1 1 0 e τ dτ = 4(1 e 1 ) 3 t < 4 0 y(t) = 2 e τ dτ + 2 t 3 1 0 e τ dτ = 4 2e 1 2e t 3 1 y(t) = 2 e τ dτ = 2(e 3 t e 1 ) t 3 8
elsewhere, y(t) = 0 Therefore, finally we have 0 t < 1 2(e t e 1 ) 1 t < 0 4 2(e t + e 1 ) 0 t < 1 y(t) = 4(1 e 1 ) 1 t < 2 4 2(e 1 + e t 3 ) 2 t < 3 2(e 3 t e 1 ) 3 t < 4 0 t 4 8. x(t) and h(t) are shown in the figure below. 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 1 < 0 results in no overlap and hence the integral evaluates to zeros i.e. y(t) = 0 If t 1 > 0 and t 2 0 i.e. 1 < t 2 9
y(t) = t 1 If t 2 > 0 and t 3 0 i.e. 2 < t 3 0 ( 2)e τ dτ = 2(e (t 1) 1) y(t) = 2 t 2 If t 3 0 i.e. t 3 e τ dτ 2 t 1 0 t 2 e τ dτ = 2(1 + e (t 1) 2e (t 2) ) 10
y(t) = 2 t 2 e τ dτ 2 t 1 t 3 t 2 e τ dτ = 2(e (t 1) 2e (t 2) + e (t 3) ) Therefore, finally we have 0 t < 1 2(e (t 1) 1) 1 t 2 y(t) = 2(1 + e (t 1) 2e (t 2) ) 2 t < 3 2(e (t 1) 2e (t 2) + e (t 3) ) t > 3 11