Unformatted text preview: of the o utput of the first lowpass filter are all shown in the figures below. Clearly this system does not accomplish the filtering task.
21. ( a) T he Nyquist r ate for t he given signal is 2 x 500011" = 1000011". Therefore, in order to be able to recover x (t) from xp(t), t he sampling period must a t most be T max = 1~~01r = 2 X 1 0 4 sec. Since the sampling period used is T = 1 0 4 < Tmax , x (t) can be recovered from xp(t). , ( b) T he Nyquist rate for the given signal is 2 x 1500011" = 3000011". Therefore, in order to be able to recover x (t) from xp(t), t he sampling period must a t most be Tmax = 30~~O'" = 0.66 X 1 0 4 sec. Since the sampling period used is T = 1 0 4 > T max , x (t) c annot be recovered from x p ( t) . ( c) Here, I m{X(jw)} is not specified. Therefore, t he Nyquist rate for the signal x (t) is indeterminate. This implies t hat one cannot guarantee t hat x (t) would be recoverable from xp(t). ( d) Since x (t) is real, we may conclude t hat X (jw) = 0 for Iwl > 5000. Therefore, the answer t o t his p art is identical to t hat o f p art ( * ( e) Since x (t) is real, X (jw) = 0 for identical to t hat o f p art (b). Iwl > 1500011". Therefore, the answer to this p art is ( f) I f X(jw) = 0 for Iwl > W I, t hen X (jw)*X(jw) = 0 for Iwl > 2w1. Therefore, in this p art, X (jw) = 0 for Iwl > 750011". T he Nyquist r ate for this signal is 2 x 750011" = 1500011". Therefore, in order to be able to recover x (t) from xp(t), t he sampling period must a t most be Tmax = 156~o". = 1.33 x 1 0 4 sec. Since the sampling period used is T = 1 0 4 < T max , x (t) can be recovered from xp(t). ( g) I f IX(jw)/ = 0 for w > 500011", t hen X (jw) = 0 for w > 500011". Therefore, the answer to this p art is identical to the answer of p art (a).
22. Using the properties o f t he Fourier transform, we o btain Therefore, Y (jw) = 0 for Iwl > 100011". T his implies t hat t he Nyquist rate for y(t) is 2 X 100011" = 200011". Therefore, the sampling period T can a t most be 211"/ (200011") = 1 0 3 sec. Therefore we have to use T < 1 0 3 sec in order to be able to recover y(t) from Yp(t). 23. ( a) We may express p(t) as p(t) = P l(t)  pt{t where pt{t) = ~), I : 6 (t k2~). Now,
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) < : to _fs'O N\?~I " EE 301 Homework 12 Solutions
1. (Problem 7.21  a,b,f,g) Mark Wilde November 27, 2005 For all these questions, simply use the Nyquist criterion from the sampling theorem, i.e., π >T ωM where ωM is the maximum angular frequency of a band limited signal. (a) ωM = 5000π . It is true that 1 1 > 5000 10000 so we can recover the original signal from the sampled version. (b) ωM = 15000π . It is not true that 1 1 > 15000 10000 so we can’t recover the original signal from the sampled version. (c) (d) (e) (f) If the convolution has maximum angular frequency ωM = 15000π , then the original signal must have half that width. So for the original signal, ωM = 7500π . It is true that 1 1 > 7500 10000 so we can recover the original signal perfectly. (g) There may be a typo in this problem. If we take it as it is, then, in general, the signal has inﬁnite bandwidth and cannot be recovered exactly. If there is a typo where it should be ω  > 5000π , then the signal is bandlimited with maximum frequency ωM = 5000π , and it reduces to problem a, since X (jω ) = 0 ⇒ X (jω ) = 0 (5) (4) (3) (2) (1) 2. Convolution in the time domain is multiplication in the frequency domain. Therefore, the 1 maximum frequency of Y (jω ) must be 1000π . So we just need a sampling period T < 1000 . 3. We just need to show that xr (t) = x(t) at the sampling instants, i.e., xr (kT ) = x(kT ) ∀ k ∈ Z To do this, plug in t = kT to ﬁgure out the sampling instants. (6) EE 301 Homework 12 Solutions Mark Wilde November 27, 2005 ∞ xr (kT ) =
n=−∞ ∞ x(nT ) x(nT )
n=−∞ π sin( T (kT − nT )) π T (kT − nT ) (7) (8) =
sin(π (k −n)) π (k −n) . sin(π (k − n)) π (k − n) Now let’s consider the function When k = n, let’s say that k − n = m and we have the following sin(πm) πm The denominator is nonzero, and an integer multiple of π makes sin(πm) = 0. When k = n, this function is equal to 1 because we consider the limit as the argument goes to zero and use L’Hospital’s Rule. Therefore, the function must be equal to the discrete delta function.
∞ (9) xr (kT ) =
n=−∞ x(nT )δ[k − n] (10) (11) = x(kT ) 4. We need to do all analysis in the frequency domain. Taking Fourier transforms, X (jω ) = F {A + B cos((2π/T )t + θ )} B = F {A} + F ej ((2π/T )t+θ) + e−j ((2π/T )t+θ) 2 B B = 2Aπδ(ω ) + ejθ F ej (2π/T )t + e−jθ F e−j (2π/T )t 2 2 B B jθ = 2Aπδ(ω ) + e 2πδ(ω − 2π/T ) + e−jθ 2πδ(ω + 2π/T ) 2 2 (12) (13) (14) (15) Since the actual values of the coeﬃcients are irrelevant, let’s just use the following variables instead. A B C So, = 2Aπ = Be π = Be
−jθ jθ (16) (17) π (18) Page 2 ...
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 Fall '08
 Enright
 Digital Signal Processing, Harry Nyquist, sampling period, Nyquist–Shannon sampling theorem, IWL, Bandlimited

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