SAMPLE SOLUTIONS DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters Andreas Antoniou

SAMPLE SOLUTIONS DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters Andreas Antoniou (Revision date: February 7, 7) SA. A periodic signal can be represented by the equation x(t) k A k sin(ω k t + φ k ) the frequencies ω k, amplitudes A k, and phase angles φ k are given in Table SA.. The signal is to be processed first by an ideal bandpass filter and the by a differentiator, as shown in Fig. SA.. The bandpass filter will pass frequencies in the range ω 6 rad/s and reject all other frequencies and the differentiator will differentiate the signal with respect to time. (a) Obtain a time-domain representation for the signal at the outputs of the bandpass filter and differentiator, i.e., at nodes B and C, respectively, in Fig. SA.. (b) Obtain a frequency-domain representation for the signal at the output of the bandpass filter. (c) Obtain a frequency-domain representation for the signal at the output of the differentiator. A Bandpass Filter B Differentiator C Figure SA. Table SA. Frequency Spectrum k ω k, rad/s A k φ k.89.78.6.8.8575.5.69.9 5 5..69 6 6.868.668 7 7.555.7 8 8.6797.778 9 9.7.58.. Copyright c by Andreas Antoniou 5-

DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters (a) NodeB: 6 x B (t) A k sin(ω k t + φ k ) k Node C: [ 6 ] x C (t) d A k sin(ω k t + φ k ) dt k 6 ω k A k cos(ω k t + φ k ) k 6 k d dt [A k sin(ω k t + φ k )] 6 ω k A k sin(ω k t + φ k + π/) k (b) The amplitude and phase spectrums at Node B are given in Table SA. and are plotted in Fig. SA.. Table SA. Frequency Spectrum at Node B k ω k, rad/s A k φ k.69.9 5 5..69 6 6.868.668.9 Amplitude Spectrum Phase Spectrum Magnitude.8.7.6.5.... Phase Angle, rads.....5.6 5 Frequency, rad/s.7 5 Frequency, rad/s Figure SA. (c) Similarly, the amplitude and phase spectrums for Node C are given in Table SA. and areplottedinfig.sa.. Table SA. Frequency Spectrum at Node B k ω k, rad/s A k φ k.56.6 5 5.67.5 6 6 5.888.56

SAMPLE SOLUTIONS 6 Amplitude Spectrum.5 Phase Spectrum 5 Magnitude Phase Angle, rads.5.5 5 Frequency, rad/s 5 Frequency, rad/s Figure SA. SA. A periodic signal defined by x(t) x(t + rτ ) r x(t) is zero outside the range τ / t τ / has a Fourier series of the form x(t) X k e jkω t k for τ / t τ / X k τ / x(t)e jkωt dt τ τ / (a) Assuming that x(t) is an even function, show that X k τ / x(t)coskω tdt τ Justify your steps. (b) The periodic signal of Fig. SA. is described by the equation { cos ω t/ for τ / t τ / x(t) otherwise ω π/τ. Using the formula in part (a), obtain an expression for the Fourier series coefficients X k. (c) Give expressions for the amplitude and phase spectrums of x(t).

DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters ~ x(t) cos(ω t/) τ τ τ τ τ t Figure SA. (a) From the definition of the Fourier series X k τ / x(t)e jkωt dt τ τ / τ / x(t)(cos kω t j sin kω t) dt τ τ / τ / x(t)coskω tdt j τ / x(t)sinkω tdt τ τ / τ τ / [ ] τ / x(t)coskω tdt+ x(t)coskω tdt τ τ / [ ] τ / x(t)sinkω tdt+ x(t)sinkω tdt τ τ / If x(t) iseven,thenx(t)coskω t is an even function and x(t)sinkω t is an odd function. Hence τ / x(t)coskω tdt x(t)coskω tdt τ / and τ / x(t)sinkω tdt x(t)sinkω tdt τ / Therefore X k τ / x(t)coskω tdt τ (b) The given signal is symmetrical about the y axis and, therefore,it is an even function. Hencewehave X k τ / x(t)coskω tdt τ τ / cos(ω t/) cos(kω t) dt τ / cos(kω t)cos(ω t/) dt τ τ From trigonometry, we have cos A cos B [cos(a + B)+cos(A B)]

SAMPLE SOLUTIONS 5 and, therefore, X k τ / τ [cos(kω t + ω t/) + cos(kω t ω t/)] dt τ / cos [( k + τ ) ω t ] +cos [( k ) ω t ] dt [ [( ) sin k + ω t ] ( ) τ k + + sin [( ) k ω t ] ] τ/ ( ) ω k ω [ (k sin ) ] [ + π (k ) ] τ τ sin π τ τ ( ) τ k + π + ( ) τ k π τ [ [( sin k + π ] ( ) ) π k + + sin [( k π ]] ( ) ) k The amplitude and phase spectrums of x(t) are the magnitude and angle of X k, i.e., X k and arg X k. Since X k is real, the angle of X k is or π depending on whether X k is positive or negative. SA. A z transform is given by X(z) (z +)(z +) (z + z )(z ) (a) Construct the zero-pole plot of X(z). (b) Function X(z) is known to have as many Laurent series as there are annuli of convergence but only one of these series is a z transform that satisfies the absolute convergence theorem (Theorem.). Identify the annulus of convergence of that series on the zero-pole plot obtained in part (a). (c) Through the use of partial fractions obtain a closed-form expression for x(nt )Z X(z). (a) X(z) can be expressed as X(z) (z +)(z +) (z + z )(z ) (z + j)(z j)(z +) (z )(z +)(z ) Hence X(z) has zeros at z ±j, andpolesatz,,. The zero-pole plot is depicted in Fig. SA.5. (b) The correct annulus is the outer annulus which can be represented by See Fig. SA.5. (c) Using Technique I, we can write z <R R X(z) z (z +)(z +) z(z )(z +)(z ) R z + R z + R z + + R z Since the poles are simple, we have (SA.) R lim z (z +)(z +) (z )(z +)(z ) () () 6

6 DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters z plane j R j Figure SA.5 R (z +)(z +) R lim z z(z +)(z ) () (z +)(z +) lim z z(z )(z ) 5 () () () (5) 6 From Eq. (SA.), we can write (z +)(z +) R lim z z(z )(z +) 5 X(z) R + R z z + R z z + + R z z 6 z z + 6 z z + + z z Therefore, for n, the use of Table. gives x(nt ) 6 δ(nt ) u(nt )+ 6 u(nt )()n + u(nt )()n 6 δ(nt )+u(nt )[ + 6 ()n + ()n ] Since the numerator degree of X(z) is equal to the denominator degree, it follows from the corollary of the initial-value theorem (Theorem.8) that x(nt ) for n<, i.e., the above solution applies for all values of n. An alternative but equivalent solution can be readily obtained by using Technique II (see p. 5) by we expand X(z) instead of X(z)/z into partial fractions. We can write X(z) (z +)(z +) (z )(z +)(z ) R + R z + R z + + R z

SAMPLE SOLUTIONS 7 Thus (z +)(z +) R lim z (z )(z +)(z ) lim z z z R (z +)(z +) R lim z (z +)(z ) () (z +)(z +) lim z (z )(z ) 5 () () (5) (z +)(z +) R lim z (z )(z +) 5 Thus and for n, we have X(z) z z + + z x(nt )δ(nt ) u(nt T ) u(nt T )()n +u(nt T )() n δ(nt )+u(nt T )[ ()n +() n ] SA. (a) Findthez transform of the following discrete-time signal for n< x(nt ) for n 5 +(n 5)T for n>5 (b) Thez transform of a discrete-time signal x(nt )isgivenby X(z) z(z z +) (z +)(z ) Using the initial-value theorem (Theorem.8), show that x(nt ) forn<. Then find x(nt )forn using the general inversion formula x(nt ) X(z)z n dz. πj Γ (a) The signal can be expressed as Hence x(nt )u(nt )+r(nt 5T ) Zx(nT )Zu(nT )+Zr(nT 5T ) Zu(nT )+z 5 Zr(nT ) z z Tz (z ) (a) The first nonzero value of x(nt ) occurs at KT (N M)T N is the denominator degree and M in the numerator degree in X(z). Since N M,wehaveK, i.e., the signal starts at KT. Hence for n<, we have x(nt )

8 DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters (b) Forn, we can write x(nt ) M i M i M i [ Res X(z)z n ] zp i Res zp i Res zp i z(z z +)z n (z +)(z ) (z z +)z n (z j)(z + j)(z ) R z j + R z + j + R z (SA.) M is the number of poles and R is the complex conjugate of R since the pole at z j e jπ/ is the complex conjugate of the pole at z j e jπ/. Since the poles are simple, we have [ (z z +)z n ] R lim (z j) z j (z j)(z + j)(z ) (z z +)z n lim z j (z + j)(z ) j j jn e jnπ/ ( j +)jn j(j ) and (z z +)z n R lim z (z +) Therefore, from Eq. (SA.), we can write x(nt )u(nt )e jnπ/ + u(nt )e jnπ/ + u(nt ) u(nt )( cos nπ/+) SA.5 An initially relaxed discrete-time system can be represented by the equation y(nt )Rx(nT ).5x(nT )+ e.(nt +T ) x(nt T )+x(nt T ) By using appropriate tests, check the system for (a) linearity, (b) time invariance, and (c) causality. (a) Linearity R[αx (nt )+βx (nt )].5[αx (nt )+βx (nt )] + e.(nt +T [αx (nt T ) +βx (nt T )] + [αx (nt T )+βx (nt T )] α[.5x (nt )+ e.(nt +T ) x (nt T )+x (nt T )] +β[.5x (nt )+ e.(nt +T ) x (nt T )+x (nt T )] αrx (nt )+βrx (nt ) Therefore, the system is linear.

SAMPLE SOLUTIONS 9 (b) Time invariance The response to a delayed excitation is Rx(nT kt).5x(nt kt)+ e.(nt +T ) x(nt kt T )+x(nt kt T ) The delayed response is y(nt kt).5x(nt kt)+ e.(nt kt+t ) x(nt kt T )+x(nt kt T ) For any k,wehave e.nt +T e.(nt kt+t ) Thus y(nt kt) Rx(nT kt) and, therefore, the system is time dependent. (c) Letx (nt )andx (nt ) be arbitrary discrete-time signals such that x (nt )x (nt ) x (nt ) x (nt ) for n k for n>k We have Rx (nt ).5x (nt )+ e.(nt +T ) x (nt T )+x (nt T ) (SA.) and Rx (nt ).5x (nt )+ e.(nt +T ) x (nt T )+x (nt T ) (SA.) Since x (nt )x (nt ) for n k then x (nt T )x (nt T ) for n k and x (nt T )x (nt T ) for n k Hence the right-hand side in Eq. (SA.) is equal to the right-hand side in Eq. (SA.) for n k and thus Rx (nt )Rx nt ) for n k Therefore, the filter is causal. SA.6 (a) Derive a state-space representation for the filter shown in Fig. SA.6. (b) Using the state-space representation obtained in part (a), compute the impulse response of the filter at nt 5T. x(nt) y(nt)..6 5 Figure SA.6

DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters x(nt) q (nt+t) q (nt+t) y(nt) q (nt) q (nt)..6 5 Figure SA.7 (a) State variables can be assigned as shown in Fig. SA.7. Henc ewecanwrite q (nt + T ) x(nt ).q (nt ) (SA.5) q (nt + T ) q (nt + T )+q (nt ).6q (nt ) (SA.6) On eliminating q (nt + T ) in Eq. (SA.6) using Eq. (SA.5), we get q (nt + T ) x(nt ).8q (nt )+q (nt ).6q (nt ).q (nt ).6q (nt )+x(nt ) (SA.7) From Eqs. (SA.5) and (SA.7) [ ] q (nt + T ) q (nt + T ) [...6 ][ ] q (nt ) q (nt ) The output is given by y(nt )q (nt + T )+5q (nt ) and from Eqs. (SA.6) and (SA.9), we have [ + x(nt ) ] y(nt ).q (nt ).6q (nt )+ x(nt )+5q (nt ) 8.8q (nt )+.6q (nt )+8x(nT ) [ ] q (nt ) [8.8.6] +8x(nT ) q (nt ) (SA.8) (SA.9) (SA.) Therefore, Eqs. (SA.8) and (SA.) can be written as q(nt + T )Aq(nT )+bx(nt ) y(nt )c T q(nt )dx(nt ) A [ ]...6 c T [8.8.6] d 8 [ b ] (b) The impulse response is given by h(nt ) { a for n c T A n b otherwise

SAMPLE SOLUTIONS For n 5 Since [ [ ] h(5t )c T A. b [8.8.6]..6] [ ] [ ][ ] [ ]....6...6..6..6..6 [ ] [ ][ ] [ ]..6..6..56...6..6..6..96 we get h(5t )[8.8.6] SA.7 Fig. SA.8 shows a recursive digital filter. [ ] [ ].56..56 [8.8.6].75..96][.888 (a) Find its transfer function. (b) By using the Jury-Marden stability criterion, determine whether the filter is stable or unstable. x(nt) y(nt) 5 Figure SA.8 (a) From Fig. SA.8, we get y(nt )x(nt ) y(nt T ) y(nt T ) y(nt T ) 5y(nT T ) Hence the z transform gives Y (z) X(z) z Y (z) z Y (z) z Y (z) 5 z Y (z) or and so In effect, Y (z)+ z Y (z)+ z Y (z)+ z Y (z)+ 5 z Y (z) X(z) Y (z) X(z) + z + z + z + 5 z H(z) N(z) D(z) z z + z + z + z + 5 We note that D(z) z + z + z + z + 5 D() + + + + 5.8 > () D() + + 5.78 > (SA.a) (SA.b)

DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters (b) The Jury-Marden array can be constructed as shown in Table SA.. Table SA. The Jury-Marden array b b b b b 5 5 5 9 5 5 9 5 5.899.9.885 From Eqs. (SA.a) and (SA.b) and Table SA., we have D() > () D() > b > 5 b c 5 > c d.899 >.885 d Therefore, conditions (i) to (iii) of the Jury-Marden stability criterion (see p. ) are satisfied and the filter is stable. SA.8 The filter of Fig. SA.9 is subjected to an input { for n andn x(nt ) otherwise Find the time-domain response in closed form if m and m 8. The filter is linear and time-invariant. x(nt) y(nt) m m Figure SA.9 From Fig. SA.9 Y (z) X(z)+z X(z)+z X(z)+m z Y (z)+m z Y (z) Hence Y (z)( m z m z )(+z + z )X(z)

SAMPLE SOLUTIONS or Therefore, Y (z) ( + z + z ) ( m z m z ) X(z) Y (z) X(z) H(z) z +z + z m z m z +z + z + z + 8 z +z + ( )( ) z + z + The time-domain response can be obtained in a number of ways, as detailed below. Method : The input is the sum of two impulse functions, i.e., x(nt )δ(nt )+δ(nt T ) Hence Rx(nT )R[δ(nT )+δ(nt T )] Since the filter is linear, we have and if h(nt ) is the impulse response, i.e., Rx(nT )Rδ(nT )+Rδ(nT T ) h(nt )Rδ(nT ) then by virtue of the fact that the filter is time-invariant, we get y(nt )h(nt )+h(nt T ) In effect, all we have to do is find the impulse response. Expanding H(z)/z into partial fractions gives H(z) z +z + z z ( ) z + )( z + R z + R z + + R z + R z +z + ( )( ) z + z + z 8 i.e., and R z +z + z ( ) z + R z +z + z ( ) z + z z ( )( + + ) H(z) R + R z z + 6 ( )( + + ) + R z z + ( 6 + 8 6 6 )( ) 9 Now Hence h(nt )R δ(nt )+ [ ( ) R n ( + R n ] ) u(nt ) 8δ(nT )+ [ ( ) n ( 9 n ] ) u(nt ) y(nt )h(nt )+h(nt T ) y(nt )8[δ(nT )+δ(nt T )] + [ ( ) n ( 9 n ] ) u(nt ) [ + ( ) n ( ) ] 9 n u(nt T )

DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters Method Since x(nt )δ(nt )+δ(nt T ) the z transform gives X(z) +z z + z Hence Y (z) H(z)X(z) (z +z +) (z +) ( )( ) z + z + z Expanding H(z)X(z) into partial fractions, gives Therefore, R lim H(z)X(z) z R (z +z +)(z +) ) ( z + H(z)X(z) R + R z + R z + )( z + R (z +z +)(z +) ( z + ) z z R (z +z +)(z +) ( z + ) z z 9 6 6 7 + R z + z 8 ( +)( ( +) ( )( ) + )( ) ( )( ) ( 6 +)( ( +) ( + )( ) 6 + )( 8 ( 6 ) )( ) y(nt )R δ(nt )+R δ(nt T )+R u(nt T ) ( ) n + R u(nt T ) ( δ(nt )+8δ(nT T )+u(nt T ) ( ) n 7 u(nt T ) ( Method The inverse of Y (z) is obtained from first principles as y(nt ) Y (z) res ) n ) n Y (z) Y (z)z n H(z)X(z)z n (z +z +)(z +) z ( ) z + )( z + z n However, watch out for pitfalls at the origin. In this case, we have a second-order pole at the origin if n, a first-order pole at the origin if n, and no poles at the origin if n. Hence, we have to find y() and y(t ) individually and then proceed to y(nt )forn. This would make this method quite long. Method We can express Y (z) into partial fractions as Y (z) (z +z +)(z +) z ( ) z + )( z + R + R z + R z z + + R z z +

SAMPLE SOLUTIONS 5 R lim zy (z) z(z +z +)(z +) z z ( ) z + )( z + z 8 z + R lim Y (z) (z +z +)(z +) z z z ( ) z + z [ ( ) ( ) ] + ( + +) ( ( ) ( + ) +)( ) ( )( ) R z + lim z z [ ( Y (z) (z +z +)(z +) z ( ) z + z ) ( ) ] + ( + +) ( 6 ( ) ( + ) +)( ) 6 Constant A can be obtained by noting that 8 6 9 6 6 7 Thus Hence Therefore, lim Y (z) R + R + R z z z R R R () 7 7 Y (z) + 8 z z z + + 7z z + y(nt )δ(nt )+8δ(nT T )+ [ 7 ( ) n ( ) n ] u(nt ) Method 5 One could expand Y (z)/z into partial fractions as Y (z) z R z + R z + R z + However, this is essentially the same as method. SA.9 A discrete-time system has a transfer function H(z) + R z + z + z (r cos θ)z + r (a) Find the unit-step response in closed form. (b) Using MATLAB, D-Filter, or similar software, plot the unit-step response for r.and θ π/. (c) Repeat part (b) forr.6 andθ π/. (d) Repeat part (b) forr.9 andθ π/. (e) Compare the unit-step responses in parts (b) to(d). (a) Thez transform of the output of the system is given by Y (z) H(z)X(z) (SA.)

6 DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters H(z) is the transfer function and X(z) is the z transform of the input. Since the input is a unit step, we have Thus Eqs. (SA.) and (SA.) give Y (z) X(z) Zu(nT ) z z z + z (r cos θ)z + r z z z +z (z )[z (r cos θ)z + r ] i (SA.) The general inversion formula gives y(nt ) M Y (z)z n dz Res Y (z) (SA.) πj Γ zp i Y (z) Y (z)z n (z +z)z n (z )[z (r cos θ)z + r ] (z +z)z n (z )[z r(e jθ + e jθ )z + r ] (z +)z n (z )(z re jθ )(z re jθ ) and M. The residues of Y (z) can be obtained as follows: (z +)z n R lim(z )Y (z) z z (r cos θ)z + r R (r cos θ)+r lim (z re jθ )Y (z) zre jθ z (z +)z n (z )(z re jθ ) zre jθ (SA.5) (r e jθ +)r n e jnθ (re jθ )(re jθ re jθ ) (r e jθ +)r n e jnθ (re jθ )j sin θ (r cos θ ++jr sin θ)r n e jnθ (r cos θ +jr sin θ)j sin θ (r cos θ ++jr sin θ)r n e jnθ [(r cos θ ) + jr sin θ]e jπ/ sin θ Me jφ r n e jnθ Mr n e j(nθ+φ) (SA.6) and Similarly, M r cos θ ++jr sin θ [j(r cos θ ) r sin θ] sin θ (r cos θ +) +(r sin θ) [(r cos θ ) +(r sin θ) ]sin θ φ tan r sin θ (r cos θ ) r tan cos θ + r sin θ tan r sin θ r cos θ + r sin θ tan (r cos θ ) π R R (SA.7)

SAMPLE SOLUTIONS 7 since complex conjugate poles give complex conjugate residues. We note that there is no additional pole at the origin when n and hence for n Eq. (SA.) gives y(nt )R + R + R (SA.8) Since the numerator degree in Y (z) does not exceed the denominator degree, we have y(nt ) for n< Therefore, for any n, Eqs. (SA.5) (SA.8) give y(nt )u(nt ) [R + Mr n e j(nθ+φ) + Mr n e j(nθ+φ)] u(nt ) [ R +Mr n cos(nθ + φ) ] (b) and(d) The step responses for the three cases are illustrated in Fig. SA.a to c. (e) On the basis of the step responses obtained, the system in part (b) is what they call overdamped, the one in part (c) iscritically damped, and the one in part (d) is referred to as underdamped. SA. A second-order digital filter has zeros z e jπ/ and z e jπ/ and poles p.5e jπ/ and p.5e jπ/ and its multiplier constant is. (a) Obtain the transfer function of the filter. (b) Obtain an expression for the gain. (c) Assuming that the sampling frequency is π, calculate the gain at ω, π/, π/, and π. (a) Since we have the zeros, poles, and multiplier constant of the filter, the transfer function can be readily constructed as H(z) (z ejπ/ )(z e jπ/ ) (z ejπ/ )(z ejπ/ ) [z (e jπ/ + e jπ/ )z +] z (ejπ/ + e jπ/ )+ [z (cos π/)z +] z (cos π/)z + [z z +] z z + (b) The gain is given by M(ω) H(e jωt [e jωt e jωt +] ) e jωt ejωt + cos ωt + j sin ωt cos ωt j sin ωt + cos ωt + j sin ωt (cos ωt + j sin ωt)+ (cos ωt cos ωt +) +(sinωt sin ωt) (cos ωt cos ωt + ) +(sinωt sin ωt) (c) Sinceω s π, wehaveπf s π/t π. Hence T s. For the frequencies given, the numerical values in Table SA.5 can be readily calculated. Table SA.5 ω cos ωt sin ωt cos ωt sin ωt π π π

8 DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters Step Response 8 6 y(nt) 5 5 5 nt (a) Step Response 8 6 y(nt) 5 nt 5 5 (b) Step Response 8 6 y(nt) 5 nt 5 5 (c) Figure SA.

SAMPLE SOLUTIONS 9 Thus ( +) +( ) M() ( + ) +( ) 5.68 ( π ) M ( +) +( ) ( + ) +( ) ( ) +( ) ( ) ( ) + ( ) +.89 ( π ) M ( +) +( ) ( + ) +( ) + ( ) + (++) +( ) M(π) ( + + ) +( ) 6 ( + +) + +.658 SA. A digital filter characterized by the transfer function H(z) (z z +) z z + and a practical D/A converter are connected in cascade as shown in Fig. SA.a. The output waveform of the D/A converter is of the form illustrated in Fig. Sb τ.t sand T is the sampling period. The sampling frequency is π rad/s. (a) Obtain an expression for the gain of just the digital filter. (b) Obtain an expression for the overall gain of the digital filter in cascade with the D/A converter. (c) Calculate the gain of just the digital filter at ω,π/, π/, and π. (d) Calculate the overall gain of the digital filter in cascade with the D/A converter at ω, π/, π/, and π. (e) Sketch (i) the gain of just the digital filter and (ii) the overall gain of the digital filter in cascade with the D/A converter, and explain the effect of the D/A converter on the amplitude response. (a) The gain is given by M(ω) H(e jωt [e jωt e jωt +] ) e jωt ejωt + cos ωt + j sin ωt cos ωt j sin ωt + cos ωt + j sin ωt (cos ωt + j sin ωt)+ (cos ωt cos ωt +) +(sinωt sin ωt) (cos ωt cos ωt + ) +(sinωt sin ωt)

DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters x(nt) Digital Filter y(nt) Practical D/A converter y(t) (a) y(t) y(kt) t kt (b) Figure SA. (b) The practical D/A converter has a gain H p (jω) τ sin(ωτ/) ωτ/ (SA.9) τ.t and T π/ω s π/π s (see Eq. (6.6) in textbook). Hence the overall gain of the digital filter in cascade with the D/A converter is given by (cos ωt cos ωt +) +(sinωt sin ωt) M T (ω) (cos ωt cos ωt + ) +(sinωt τ sin ωτ/ sin ωt) ωτ/ (c) Sinceω s π, wehaveπf s π/t π. Hence T s. For the frequencies given, the numerical values in Table SA.6 can be readily calculated. Table SA 6 ω cos ωt sin ωt cos ωt sin ωt π π π

SAMPLE SOLUTIONS Thus ( +) +( ) M() ( + ) +( ) 5.68 ( π ) M ( +) +( ) ( + ) +( ) ( ) +( ) ( ) ( ) + ( ) +.89 ( π ) M ( +) +( ) ( + ) +( ) + ( ) + (++) +( ) M(π) ( + + ) +( ) 6 ( + +) + +.657 (d) SinceT s, Eq. (SA.9) gives H P (jω).t sin(.tω/).tω/. sin(.ω/).ω/ p. Hence the overall gain of the digital filter in cascade with the D/A converter is obtained from the above numerical values as follows: SA. Realize the transfer function ( H(z) M T ().68..68 ( π ) M T.89..89 ( π ) M T.. M T (π).557..557 z z.5+j. + ) ( z z z.5 j. z +. + using two second-order canonic filter sections in cascade. The transfer function can be expressed as H(z) z.5z.5z z z +.5 +. z +.5z +5z +z z +.9z +. z z z z +. 8z +.5z z +.9z +. z z +.z 8+.5z +.9z +.z H (z) H (z) 5z ) z +.5

DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters H (z) z 8+.5z z +.z and H (z) +.9z +.z Now if we realize H (z) andh (z) in terms of direct canonic sections, the cascade realization shown in Fig. SA. can be readily obtained.. H (z) 8.9.5. H (z) Figure SA. SA. An analog elliptic lowpass filter with a cutoff frequency of rad/s has a transfer function of the form.75(s +.6) H(s) (s +.8)(s +.s +.5) (a) By applying the lowpass-to-highpass transformation s s get a continuous-time highpass transfer function. (b) Construct the zero-pole plot of the continuous-time highpass transfer function. (c) Using the zeros and poles obtained in part (b), get a corresponding discrete-time highpass transfer function using the matched-z-transformation method. The sampling frequency is ω s rad/s.

SAMPLE SOLUTIONS (d) How does the matched-z-transformation method compare with the invariant-impulseresponse method? The transfer function has zeros at and poles at z z, z z +j.65 z p, p,p p.8 + j. p.55 + j.697 (a) The highpass transfer function is obtained as H HP ( s) H(s).75( s +.6) (s +.8)(s +.s +.5) s s.75( s +.6) ( s +.8)( s +. s +.5).75 s( +.6 s ) ( +.8 s)(.5 s +. s +).75.6.8.5 s( s +.6 ) ( s +.8 )( s +..5 s +.6 Therefore, the highpass filter has zeros at and poles at.5 ) s/ s s( s +.86) ( s +.66)( s +.679 s +.969) s s, s, s s, s, s ± j.6 s p, p, p p.66, p, p. ± j.669 (b) The transfer function of the digital filter is given by L for a highpass filter. Hence M H (z e s it ) H D (z) (z +) L i N (z e p it ) H D (z) i (z z i ) i (z p i ) i

DIGITAL SIGNAL PROCESSING: Signals, Systems, and Filters z e T, p e.66t, z, z e ±j.6t p, p e (.±j.669)t with T π π ω s π (c) The method works with all types of filters, i.e., LP, HP, BP, and BS, and is easy to apply. However, it tends to increase the passband ripple. SA. A lowpass analog filter has a transfer function H A (s) s + s + (a) Assuming a sampling frequency of π, design a digital filter using the bilinear transformation method. (b) Find the -db and -db frequencies of the analog filter. (c) Find the -db and -db frequencies of the digital filter. (d) What should be the -db frequency of the analog filter to get a -db frequency in the digital filter at rad/s? (a) The sampling period is give by T f s π ω s π π 5 The discrete-time transfer function is given by H D (z) H A (s) s z T z+ s + s + [ (z) z+ s (z) z+ ] [ + (z) z+ ] + z +z + (z z +)+ (z ) + (z +z +) z +z + b z + b z + b b +86.86 b + 98. b + + 5. Alternatively, by factorizing constant b in the denominator, the transfer function can be expressed as H D (z) H z +z + z + b z + b H 8.685, b.75, b.7

SAMPLE SOLUTIONS 5 (b) The frequency response of the analog filter H A (jω) ω + j ω + Hence the loss is given by A(ω) log ( ω ) +ω log( ω + ω +ω ) Thus +ω. A(ω) By letting A(ω) db, the -db frequency is obtained as ω (. ) /.7 rad/s By letting A(ω) db, the -db frequency is obtained as ω (. ) / 5.6 rad/s (c) A frequency ω i in the analog filter transforms into a frequency Ω i given by Ω i T tan ω it (SA.) Hence the - and -db frequencies in the digital filter are obtained as.7 Ω tan.7 rad/s and 5.6 Ω tan 5. rad/s respectively. (d) Now if the -db frequency in the digital filter is required to be rad/s, then according to Eq. (SA.) the -db frequency in the analog filter should be ω T tan Ω T tan. rad/s