2.161 Signal Processing: Continuous and Discrete

Transcription

1 MIT OpenCourseWare Signal Processing: Continuous and Discrete Fall 008 For information about citing these materials or our Terms of Use, visit:

2 M MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING.6 Signal Processing - Continuous and Discrete Fall Term 008 Solution of Problem Set Assigned: Oct., 008 Due: Oct. 9, 008 Problem : We have shown in class (see the Fourier handout): F {sin(ω c t)} = jπ (δ(ω ω c ) δ(ω + ω c )) F {cos(ω c t)} = π (δ(ω ω c ) + δ(ω + ω c )) (a) When f audio 0, f AM (t) = sin(ω c t), and from above F. M M? M? (b) There are several ways of doing this. For example (i) Expand f AM (t) into sinusoidal components using this trigonometric relationship: cosαsin β = (sin (α + β) + sin (β α)) f AM (t) = ( + af audio (t))sin (ω c t) = ( (0.5 cos(π 000t) cos(π 000t))sin(ω c i) = sin(ω c t) = sin(ω c t) +0.5 cos(π000t) sin(ω c t) +0.5 cos(π000t) sin(ω c t) +0.5 (sin((ω c + 000π)t) + sin((ω c 000π)t)) (sin((ω c + 000π)t) + sin((ω c 000π)t))

3 and take the Fourier transform of each of the five components: F AM (jω) = jπ{(δ(ω ω c ) δ(ω + ω c )) (δ(ω (ω c 000π)) δ(ω + (ω c 000π))) (δ(ω (ω c + 000π)) δ(ω + (ω c + 000π))) (δ(ω (ω c 000π)) δ(ω + (ω c 000π))) (δ(ω (ω c + 000π)) δ(ω + (ω c + 000π))) } giving a total of 0 impulse components in the spectrum. (ii) Alternatively you can recognize that the expansion to f AM (t) = sin(ω c t) cos(π000t) sin(ω c t) cos(π000t) sin(ω c t) involves time-domain products and these will result in frequency-domain convolutions, so that F AM (jω) = F c (jω) + F c (jω) F 000 (jω) + F c (jω) F 000 (jω) π π where F c (jω) = F {sin(ω c t)}, F 000 (jω) = F {0.5 cos(000πt)}, and F 000 (jω) = F {0.5 cos(000πt)}. The same result as in (i) will follow. H ω c 0 ω c ω 000π ω +000π c c ω 000π ω +000π c c ω c 000π ω c +000π ω c 000π ω +000π c (c) The following generalizes the results of part (b) H a a 0

4 = G (d) From the above figure it can be seen that the required bandwidth is ω u rad/s. Problem : This problem uses the time-reversal property of the Fourier transform, if F {f(t)} = F(jω) then F {f( t)} = F( jω), and if f(t) is real then F( jω) = F(jω), so that F {f( t)} = F(jω). Method(). G(jω) = F(jω)H(jω).. X(jω) = G(jω)H(jω). 3. Y (jω) = X(jω) = F(jω)H(jω)H(jω) = F(jω)H(jω)H(jω) = F(jω) H(jω). The equivalent transfer function is which is real, that is with zero phase shift. Method(). G(jω) = F(jω)H(jω).. X(jω) = F(jω)H(jω). H eq = H(jω) 3. Y (jω) = G(jω) + X(jω) = F(jω)H(jω) + F(jω)H(jω) = F(jω)R {H(jω)}. The equivalent transfer function is which is real, that is with zero phase shift. H eq = R {H(jω)} Note that because it squares the frequency response magnitude, method () will have a sharper cut-off characteristic than method (). Problem 3: Note: There was a typo in the problem statement, that stated that the phase-shift at 50 Hz should be π π. The intention was for a phase shift of rad. The all-pass transfer function will only generate a phase-lead, therefore an extra inversion is required to create a lag. No penalty is imposed for missing this point. M I F = A M : I =

5 K! = N ) + L = " O L! H(jω) = θ φ = θ π since θ + φ = π, then: ( ) ω H(jω) = π tan. a π First Solution: this solution requires three op-amps for phase shift. At 50 Hz: ( ) π 00π H(jω) = = π tan. a giving a = 00π. The filter can be achieved with the following block diagram: I N I Consider the following circuit, which is derived from Fig. 7 in the op-amp filter handout: 8 E ) L # $ )! 8 K J For amplifier A V in v v v = but v = v R R and for amplifier A v = v or v = R 3 Csv R3 Cs Eliminating v v R =. V in R R 3 Cs + (R + R ) For amplifier A 3 R6 R6 V out = v v R5 R ( ) R 6 R 3 C R 6 = s v R 5 R

6 ) + " ) L and substituting for v gives the transfer function V out (s) R R 6 s R 5 /(R 3 R C) =. V in (s) R R 5 s + (R + R )/(R R 3 C) For an all-pass filter with a = 00π we require R 5 R + R = 00π and = 00π. R 3 R C R R 3 C Let C = µf, R 3 = R = 0 kω, then R 5 = 3. kω. Let R = 0 kω, then R =. kω. For unity gain we require R R 6 R R 5 =, or R 6 = =.6 kω. R R 5 R Alternative Solution: this solution requires only two op-amps for π phase shift! Write the transfer function as: H(s) = s a a = s + a s + a and simply implement a first-order block and a summer. Consider the circuit shown below: 8 E L! # 8 K J For amplifier A, v /V in = Z in /Z f, where Z in = R and Z = R /(R Cs+) are the input and feedback impedances respectively. Then v V in = R C. (s + /(R C)) Amplifier A is simply an inverting summer, and R 5 R 5 v out = R 3 V in R v R 5 R 5 = V in +. V in R3 R R C (s + /(R C)) Let R 3 = R = R 5, then ( ) V out /(R C) =. V in (s + /(R C))

7 which actually has the implicit inversion required in the (erroneous) problem statement. Let R 3 = R = R 5 = 0 kω, and C = 0.µF. Then = a = 00π giving R = 3.83kΩ R C = a = 00π giving R = 5.9kΩ R C Problem : (a) Define Ω c as the either upper or lower -3db (0.707) response frequencies. So from the definition of Ω c : jω c H bp (jω c ) = = (jωc ) p + jω c + Ω Ω Now define α = c > 0, then: Ωp Ω c α values can be found from: α = ( + ) ± ( + ) Now define equation roots as α u > α l > 0: Ω = Ωp Ω Ω + p jω c α = ( + ) ± + ( ) α = ( + ) ± + ( ) α u = ( + ) + + ( ) c = α α α + j ( α ) = ( α ) + ( α ) + α α + α = α α ( + + = 0 )α p

8 αl = ( + + ( ) ) Now note that for a stable filter > 0 and hence: ( ) Ω u = Ω p ( ) Ω l = Ω p + + (b) Δ = (Ω u Ω l ) > 0 Δ = Ω p(α u + α l α u α l ) Δ = Ω p (( + ) ( + ( + ( ) ) )) Δ = Ω p( + ( + + ) Δ = Ω ( + p ) Δ = Ω p ( ) Δ = Δ = Alternatively, we can further simplify Ω u and Ω l by realizing that we have a complete square form under square root and proceed from there: ( ) Ω u = Ω p = Ωp ( + + ) = ( + + ) () () ( ) Ω l = Ω p + + = Ω p ( + ) () = Ω p ( + ) () Δ = (Ω u Ω l ) =

9 (c) Note that = 00 (π) rad/s and Δ = 0 (π) rad/s: H bp (s) = H bp (s) = s Ωp s + s + Ω p Δs s + Δs + Ω p H bp (s) = 0πs s + 0πs + (00π) π Problem 5: Given y(t) = sin(ω 0 t) and y (t) = sin(ω t) with ω 0 < ω N = < ω ΔT (a) Assume ω = kω N ω 0 for k a positive integer, then and when t = nδt, for integer N, π y (t) = sin((kω N ω 0 )t) = sin((k ω 0 )t) ΔT y (nδt) = sin( ω 0 NΔT) = sin(ω 0 NΔT) = y(nδt) (b) Assume ω = kω N + ω 0 for k a positive integer, then π y (t) = sin((kω N + ω 0 )t) = sin((k ΔT and when t = nδt, for integer N, + ω 0 )t) y (nδt) = sin(ω 0 NΔT) = y(nδt) (c) To graphically demonstrate the concept of frequency folding and sign changes (i) Assume ΔT =.0 s. so that ω N = 50π rad/s (5Hz). Let ω = 80π rad/s (0Hz) so that k = and ω 0 = 0π rad/s (0 Hz). The following plot shows y (t) = sin(80πt), the samples taken at 0.0 sec intervals, and y 0 (t), showing the outof-phase aliased component at a frequency of 0π rad/s.

10 y(t) = sin(80 pi t) y(t) y(ndt) (samples) y(t) (aliased) t (sec) (ii) Assume ΔT =.0 s. so that ω N = 50π rad/s (5 Hz). Let ω = 0π rad/s (60 Hz) so that k = and ω 0 = 0π rad/s (0Hz). The following plot shows y (t) = sin(0πt), the samples taken at 0.0 sec intervals, and y 0 (t), showing the in-phase aliased component at a frequency of 0π rad/s. y(t) = sin(0 pi t) y(t) y(ndt) (samples) y(t) (aliased) t (sec) The above examples show that frequencies of 60 Hz and 0 Hz will both be aliased to an apparent component of 0 Hz. (d) From parts (a) and (b) we see that an under-sampled sinusoid, that is a sinusoids is

11 above the Nyquist frequency ω N will fold into either an in-phase or out-of phase sinusoid, The frequencies of the folded sinusoids are given by: ω 0 = { ω kω N, ω > ω N (in-phase) kω N ω, ω < ω N (out-of-phase) Given y(t) = 5 sin(π(5)t) + sin(π(75)t) + 3 sin(π(5)t) sampled at 00 samples/sec. Half the sampling frequency is 50 Hz (ω N = 00π rad/s). Using the equation above we find that the 75Hz sinusoid folds into an out-of-phase sinusoid at 5 Hz, and the 5 Hz sinusoid folds into a 5 Hz in-phase sinusoid. Therefore the aliased waveform is y(t) = 5 sin(π(5)t) sin(π(5)t) + 3 sin(π(5)t) = 6 sin(π(5)t) The following plot demonstrates this. 8 6 y(t) 0 6 y(t) aliased samples t (sec)