Ch. 8 Analysis of Continuous- Time Systems by Use of the Transfer Function

8.1 Stability and the Impulse Response Consider a causal linear time-invariant continuous-time system with input x(t) and output y(t). Let the system function, H(s), be finite dimensional. The impulse response, h(t), is the inverse Laplace transform of H(s).

8.1 (cont) A system is stable if all the poles of H(s) are located in the open left-half plane. A system is marginally stable if its impulse response is bounded. A system is unstable if the impulse response grows without bound as t goes to infinity.

8.2 Routh-Huritz Stability Test There are procedures for testing for stability that do not require the computation of the poles. Let A(s) = a N s N + a N-1 S N a 1 s + a 0 A necessary (but, in general, insufficient) condition for stability is that all of the above coefficients are greater than 0.

Routh-Huritz Test Gives necessary and sufficient conditions for stability. Table 8.1 Routh Array –N+1 rows, indexed on the power of s. –(N/2) + 1 columns if N is even. –(N+1)/2 columns if N is odd. –First two rows contain coefficients as shown, starting with a N.

Routh-Hurwitz Test (cont.) The other rows are computed using the two rows above. Example 8.2: Second Order Case –Let A(s) = s 2 + a 1 s + a 0. –N = 2 is even, so there are (2/2 + 1) = 2 columns. –Result: See Table 8.2 Example 8.3: Third-Order Case –Result: See Table 8.3. Example 8.4 Higher Order Case.

8.3 Analysis of the Step Response Given a system with Y(s) = B(s)/A(s) X(s). Now when x(t) = u(t), then X(s) = 1/s. So the Laplace Transform of the step response is given by Y(s) = B(s)/A(s) (1/s). Using the residue formula –Y(s) = E(s)/A(s) + c/s –Where c = [sY(s)] s=0 = H(0) Then y(t) = y 1 (t) + H(0), t≥0 –Where y 1 (t) is the Inverse Laplace transform of E(s)/A(s). Note: y 1 (t) is the transient response and H(0) is the steady state value of the step response.

8.3.1 First Order Systems Let H(s) = k/(s-p), where k is a constant and p is a pole. Then the step response is y(t)=(-k/p)(1-e pt ) The transient response is y 1 (t)= (k/p)(e pt ) The steady state value is H(0) =(- k/p) Fig. 8.3 Step response with p = 1,2,3 Fig. 8.4 Step response with p = -1,-2,-3 Fig what is H(0)?what is p?

8.3.2 Second Order System Let H(s) = (k) / [s 2 + 2ζω n s + (ω n ) 2 ] Here we have –ζ is the damping ratio –ω n is the natural frequency The quadratic formula gives us the poles: – p 1 = -ζω n + ω n SQRT( ζ 2 -1) – p 1 = -ζω n - ω n SQRT( ζ 2 -1)

Second Order Systems Case 1: Both poles are real –y tr (t) = (k/p 1 p 2 ) (k 1 e p1t + k 2 e p2t ), t≥0 –H(0) = k/(p 1 p 2 ) Case 2: Both poles are real and repeated Case 3: Poles are complex pairs—Fig. 8.8 Comparing Parameters –Underdampled when ζ <1 –Criticallly damped when ζ = 1 –Overdampled when ζ > 1 –Damped Natural Frequency = ω d = ω n SQRT( ζ 2 -1)

8.4 Response to Sinusoids and Arbitrary Inputs Let x(t) = C cos ω o t, t≥0 Using Laplace transforms it can be shown that the output of system H(s) is y(t) = y tr (t) + y ss (t) In section 5.1 the steady state response was shown to be the input scaled by the magnitude of the frequency response evaluated at frequency of the sinusoid and with an additional phase angle equal to the phase of the frequency response evaluated at the frequency of the sinusoid.

8.4 (cont.) Let x(t) be an arbitrary input. Let X(s) = C(s)/D(s) be the Laplace transform of the input. Then if the system function is H(s)=B(s)/A(s) the Laplace transform of the output is Y(s) = B(s) C(s)/A(s)D(s) Using partial fraction expansion we have Y(s)= E(s)/A(s) and F(s)/D(s). And finally y(t) = y tr (t) + y ss (t). Note that the form of the transient response depends on the poles of the system function and the form of the steady state response depends on the poles of the input (assuming stability).

8.5 Frequency Response Function The magnitude function is sometimes given in decibels. |H(ω)| db = 20 log 10 |H(ω)| Bode Plots — plots of the magnitude and phase angle of the frequency response as a function of frequency, ω, on a logarithmic scale discusses the construction of Bode Plots.

8.5.3 Bode Plots H(s) = A(s+C 1 )…(s+C M )/s(s+B 1 )…(s+B N-1 ) Let s=jω H(ω)= A(jω+C 1 )…(jω +C M )/jω(j ω +B 1 )…(j ω+B N-1 ) Let K = AC 1 C 2 …C M /B 1 B 2 …B N-1 H(ω)= K(jω/C 1 +1)…(jω /C M +1) /jω(jω/B 1 +1)…(jω/B N-1 +1) Log(AB) = Log(A) + Log(B)

8.5.3 Bode Plots (cont.) ǀ H(ω) ǀ db = 20 log ǀ K ǀ +20 log ǀ (jω/C 1 +1) ǀ … -20log ǀ jω ǀ - 20log ǀ (jω/B 1 +1) ǀ … The phase angle is shown on page 455. Constant factors: K (jωT + 1) factors: the magnitude db can be approximated as 0 for values less than the corner frequency and an increase of 20 db per decade for frequencies greater than the corner frequency (1/T).

8.6 Causal Filters In real-time filtering, ideal filters cannot be implemented because they are not causal. The magnitude functions of the causal approximations have gradual transitions from the passband to the stopband. See figure When the peak value is normalized to 1, the 3dB point (|H(ω)| =.707) marks the passband. Usually the stopbands are down 40 or 50 dB and the region between passbands and stopbands is called the transition region.

8.6.1 Butterworth Filters Two pole lowpass Butterworth filter: H(s) = (ω n 2 ) /(s 2 + sqrt(2)ω n s + ω 2 ) Maximally flat —the magnitude variation is as small as possible. The cutoff frequency (ω n ) is the -3dB point of the magnitude function. Figures 8.33 and 8.35 illustrates the difference between one, two, and three poles.

8.6.2 Chebyshev Filters Monotone—means that the magnitude curve is gradually decreasing over the pass band or stop band. The Butterworth filter is monotone in both the pass and stop bands. Type 1 Chebyshev is monotone decreasing in the stop band, but has equal ripple in the pass band. Type 2 Chebyshev is the opposite. Figures 8.37 and 8.38 compares Butterworth and Chebyshev filters.

8.6.3 Frequency Transformation Starting with any lowpass filter having a transfer function H(s), we can modify the cutoff frequency of the filter or construct other types of filters (high-pass, etc.) If the cut-off frequency is ω c =ω 1 and we want to change it to ω 2, then H(s) we replace every s by s ω 1 /ω 2. To convert a lowpass with cut-off frequency is ω 1 to a highpass filter with a passband starting at ω 2, we replace s in H(s) with ω 1 ω 2 /s. Page 473 gives transformations for bandpass and bandstop filters.