Signals and Systems Assignment Help

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Problems P18.1 Consider the system in Figure P18.1-1 for discrete-time processing of a continuoustime signal using sampling period T, where the C/D operation is as shown in Figure P18.1-2 and the D/C operation is as shown in Figure P18.1-3.

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The filter G(Q) is the lowpass filter shown in Figure P18.1-4.

The Fourier transform of xc(t), Xc(w) is given in Figure P18.1-5.

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The sampling frequency is 8 kHz. Sketch accurately the following transforms. (a) X,(w) (b) X(Q) (c) Y(Q) (d) Ye(w) Consider the continuous-time frequency response in Figure P18.2.

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We want to implement this continuous-time filter using discrete-time processing. (a) What is the maximum value of the sampling period T required? (b) What is the required discrete-time filter G(Q) for T found in part (a)? (c) Sketch the total system.

P18.3 The system in Figure P18.3 is similar to that demonstrated in the lecture. Note that, as in the lecture, there is no anti-aliasing filter.

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P18.4 Suppose we want to design a variable-bandwidth, continuous-time filter using the structure in Figure P18.4-1.

Find, in terms of wc, the value of the sampling period To and the corresponding value co, such that the total continuous-time filter has the frequency response shown in Figure P18.4-2.

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P18.5 Consider the system in Figure P18.5-1.

Let H(Q) be as given in Figure P18.5-2 and X(co) as given in Figure P18.5-3.

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(a) Sketch X(Q) and Y(Q). (b) Suppose we replace the system in Figure P18.5-1 by the P18.5-4. Find G(w) such that y[n] = z[n].

P18.6 Suppose we are given the system in Figure P18.6-1.

(a) Find the appropriate values of the sampling period To to avoid aliasing. Also find the proper value for K so that the overall system has a gain of unity at w = 0 (i.e., no overall dc gain).

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(b) Suppose To is halved, but the anti-aliasing and reconstruction filters are not modified. (i) If X(w) is as given in Figure P18.6-2, find Y(Q).

(ii) If Y(Q) is as given in Figure P18.6-3, find Y(w).

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P18.7 Figure P18.7 shows a system that processes continuous-time signals using a digital filter. The digital filter h[n] is linear and causal with difference equation

For input signals that are bandlimited so that Xe(w) = 0 for I l > ir/T, the system is equivalent to a continuous-time LTI system. Determine the frequency response He(w) of the equivalent overall system with input xc(t) and output yc(t).

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P18.8 Figure P18.8-1 depicts a system for which the input and output are discrete-time signals. The discrete-time input x[n] is converted to a continuous-time impulse train x, (t). The continuous-time signal x,(t) is then filtered by an LTI system to produce the output yc(t), which is then converted to the discrete-time signal y[n]. The LTI system with input xc(t) and output yc(t) is causal and is characterized by the linear constantcoefficient difference equation

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The overall system is equivalent to a causal discrete-time LTI system, as indicated in Figure P18.8-2. Determine the frequency response H(Q) of the equivalent LTI system.

P18.9 We wish to design a continuous-time sinusoidal signal generator that is capable of producing sinusoidal signals at any frequency satisfying wi : W5 W2, where w, and W2 are positive numbers.

Our design is to take the following form. We have stored a discrete-time cosine wave of period N; that is, we have stored x[O], . . . , x[N - 1], where

Every T seconds we output an impulse weighted by a value of x[k], where we proceed through the values of k = 0, 1, ... , N - 1 in a cyclic fashion. That is,

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(a) Show that by adjusting T we can adjust the frequency of the cosine signal being sampled. Specifically, show that

where wo = 21r/NT. Determine a range of values for T so that y,(t) can represent samples of a cosine signal with a frequency that is variable over the full range (b) Sketch Y,(w).

The overall system for generating a continuous-time sinusoid is depicted in Figure P18.9-1. H(w) is an ideal lowpass filter with unity gain in its passband:

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The parameter we is to be determined such that y(t) is a continuous-time cosine signal in the desired frequency band. (c) Consider any value of T in the range determined in part (a). Determine the minimum value of N and some value for w,such that y(t) is a cosine signal in the range wi : O o02. (d) The amplitude of y(t) will vary depending on the value of wchosen between wi and W2. Determine the amplitude of y(t) as a function of w and as a function of N.

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P18.10 In many practical situations, a signal is recorded in the presence of an echo, which we would like to remove by appropriate processing. For example, Figure P18.10-1 illustrates a system in which a receiver receives simultaneously a signal x(t) and an echo represented by an attenuated delayed replication of x(t). Thus, the receiver output is s(t) = x(t) + ax(t - TO), where jal < 1. The receiver output is to be processed to recover x(t) by first converting to a sequence and using an appropriate digital filter h[n] as indicated in Figure P18.10-2.

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Assume that x(t) is bandlimited, i.e., X(w) = 0 for IwI > wm, and that al < 1. (a) If To < lr/M and the sampling period is taken equal to To (i.e., T = TO), determine the difference equation for the digital filter h[n] so that yc(t) is proportional to x(t). (b) With the assumptions of part (a), specify the gain A of the ideal lowpass filter so that yc(t) = x(t). (c) Now suppose that 2r/wm < To < 27r/WM. Determine a choice for the sampling period T, the lowpass filter gain A, and the frequency response for the digital filter h[n] such that yc(t) is equal to x(t).

Solutions S18.1 (a) Since x,(t) = xc(t)p(t), then X,(w) is just a replication of Xe(w) centered at multiples of the sampling frequency, namely 8 kHz or 27r8 X 10' rad/s. The sampling period is T = 1/8000.

(b) X(Q) is just a rescaling of the frequency axis, where 21r8 X 103 becomes 2 1r.

X(Q) is shown in Figure S18.1-2.

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(c) Y(Q) is the product G(Q)X(Q). Therefore, Y(Q) appears as in Figure S18.1-3.

(d) Y,(w) is a frequency-scaled version of Y(w) but only in the range 0 = --w to 7, as shown in Figure S18.1-4. Also note the gain of T.

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S18.2 (a) The maximum nonzero frequency component of H(w) is 5 00 w. Therefore, this frequency can correspond to, at most, the maximum digital frequency before folding, i.e., Q = x. From the relation wT = Q,we get

(b) Since w = 500w maps to Q = 7r, the discrete-time filter G(Q) is as shown in Figure S18.2-1.

(c) The complete system is given by Figure S18.2-2. Note the need for an antialiasing filter.

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S18.3 (a) Recall that Xc(w) is as given by Figure S18.3-1.

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Ye(w) is given by eq. (S18.3-2) and Figure S18.3-3.

Thus x(t) = y(t) in this case. (b) Xc(w) is as given in Figure S18.3-4.

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We now use eq. (S18.3-1), shown in Figure S18.3-5.

Thus, in the range ± r, X(Q) = 20000 E",,U X,[20000(9 - 2irn)] is given as in Figure S18.3-6.

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Using eq. (S18.3-2), we find Ye(w) as in Figure S18.3-7.

Note aliasing since 27000 Hz is above half the sampling rate of 20000 Hz. (c) Xe(w) is as given in Figure S18.3-8.

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Again we use eq. (S18.3-1), shown in Figure S18.3-9.

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Thus X(Q) is given as in Figure S18.3-10.

Finally, from eq. (S18.3-2) we have Y,(w) shown in Figure S18.3-11.

S18.4 It is required that we sample at a rate such that the discrete-time frequency 7r/ 2 will correspond to c. The relation between 9, and cis Q, = ocTo. Thus, we require

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As wc increases, demanding a wider filter, To decreases, and consequently the sampling frequency must be increased. There are two ways to calculate Wa. First, since we are sampling at a rate of

we need an anti-aliasing filter that will remove power at frequencies higher than half the sampling rate; therefore wa = 2wc. Alternatively, we note that the "folding frequency," or the frequency at which aliasing begins, is 0 = 7r. Since 9 = r/2 corresponds to wc, then r must correspond to 2w,.

S18.5 (a) We sketch X(Q) by stretching the frequency axis so that 2 7 corresponds to the sampling frequency with a gain of 1/TO. We then repeat the spectrum, as shown in Figure S18.5-1.

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After filtering, Y(Q) is given as in Figure S18.5-2.

(b) We see that Y(Q) looks like X(w) filtered and then sampled. The discrete-time frequency is -/3. Again, 2-x corresponds to 27r/To, so 7/3 corresponds to ir/3T. Thus, if x(t) is filtered by G(w) as given in Figure S18.5-3, then y[n] = z[n].

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Solutions to Optional Problems S18.6 (a) Since we are allowing all frequencies less than 1007r through the antialiasing filter, we need to sample at least twice 100r, or 200r. Thus, 2 007r = 2x/T or To = 10 ms. To find K, recall that impulse sampling introduces a gain of 1/TO. To account for this, K must equal To, or K = 0.01. (b) (i) Since X(w) is bandlimited to 100-r, the anti-aliasing filter has no effect. The Fourier transform of x,(t), the modulated pulse train, is given in Figure S18.6-1.

Since To = 0.005, the sampling frequency is 4007r. After conversion to a discrete-time signal, X(Q) appears as in Figure S18.6-2.

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(ii) There are three effects to note in D/C conversion: (1) a gain of To, (2) a frequency scaling by a factor of T,, and (3) the removal of repeated spectra. Thus, Y(w) is as shown in Figure S18.6-4.

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S18.7 After the initial shock, you should realize that this problem is not as difficult as it seems. If instead of h[n] we had been given the frequency response H(Q), then He(w) would be just a scaled version of H(Q) bandlimited to ir/T. Let us find, then, H(Q). Using properties of the Fourier transform, we have

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Therefore, the magnitude and phase of He(c) are as shown in Figure S18.7.

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From our previous study, we know that Xe(o) in the range ±ir/Tlooks just like X(Q) in the range ±ir. Similarly, Y,(w) between -- r/T and +r/T looks like Y(Q) in the range -ir to r. Although there is a factor of T, we can disregard it in analyzing this system because it is accounted for in the H(w) filter. The transformation of xc(t) to yc(t) will correspond to filtering x[n], yielding y[n]. In fact, the equivalent system will have a system function H(Q) given by

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where He(w) is the Fourier transform of h(t). Thus, we need to find Hc(w). The relation between yc(t) and xc(t) is governed by the following differential equation:

Using the properties of the Fourier transform, we have

S18.9 (a) It is instructive to sketch a typical y,(t), which we have done in Figure S18.9-1.

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Let us suppose that T is changed by being reduced. Then the envelope of y,(t) seems to correspond to a higher-frequency cosine. At time kt,

where we use the sampling property of the impulse function. Thus,

(b) Recall that sampling with an impulse train repeats the spectrum with a period of 2r/T and a gain factor of 1/T. Since 5([cos(2t/NT)J is as given by Figure S18.92, Y,(w) is then given by Figure S18.9-3.

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(c) The minimum value of N is 2, corresponding to the impulses at o and (27r/T ­ wo) being superimposed at w/T. The lowpass filter cutoff frequency must be such that the (superimposed) impulses at 7r/T are in the passband and those at 37r/T are outside the passband. Consequently,

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(d) Comparing Y(w) and Y,(w) in Figures S18.9-2 and S18.9-3 respectively, we see that for N > 2 the cosine output will have an amplitude of 1/T = w/21r. If N = 2, then the output amplitude will be 2/T = w/7r.

S18.10 (a) By sampling sc(t), we get

This is a first-order difference equation, so given s[n], we can find x[n]. Since x(t) is appropriately bandlimited, we can then set

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(b) From part (a) we see that T = A will make y(t) = x(t). (c) Since we do not want to alias, we still need T < 7r/wM. Now

Taking the continuousFourier transform, we see that

Thus, the continuous-time inverse system has frequency response

We want to implement this in discrete time. Therefore, using the relation, we obtain