Transfer Function and the Laplace Transformation

Added on -2020-02-05

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Q.No:1(A)The transfer function provides a basis for determining important system response characteristicswithout solving the complete differential equation. As defined, the transfer function is a rationalfunction in the complex variable s = σ + jω, that isH(s) = bmsm + bm−1sm−1 + ... + b1s + b0ansn + an−1sn−1 + ... + a1s + a0(1)It is often convenient to factor the polynomials in the numerator and denominator, and to writethe transfer function in terms of those factors:H(s) = N(s)D(s) = K (s − z1)(s − z2)...(s − zm−1)(s − zm)(s − p1)(s − p2)...(s − pn−1)(s − pn), (2)where the numerator and denominator polynomials, N(s) and D(s), have real coefficients definedby the system’s differential equation and K = bm/an. As written in Eq. (2) the zi’s are the rootsof the equationN(s)=0, (3)and are defined to be the system zeros, and the pi’s are the roots of the equationD(s)=0, (4)and are defined to be the system poles. In Eq. (2) the factors in the numerator and denominatorare written so that when s = zi the numerator N(s) = 0 and the transfer function vanishes, that is
lims→ziH(s)=0.and similarly when s = pi the denominator polynomial D(s) = 0 and the value of the transferfunction becomes unbounded,lims→piH(s) = ∞.All of the coefficients of polynomials N(s) and D(s) are real, therefore the poles and zeros mustbe either purely real, or appear in complex conjugate pairs. In general for the poles, either pi = σi,or else pi, pi+1 = σi±jωi. The existence of a single complex pole without a corresponding conjugatepole would generate complex coefficients in the polynomial D(s). Similarly, the system zeros areeither real or appear in complex conjugate pairs.(B)(i)We wish to know if a system with transfer function G(s) = 1 s + γ is stable. If γ < 0, let u(t) = σ(t), the unit step function, which is obviously bounded. Then, y(t) = L −1 [ 1 s (s + γ) ] = L −1 [ 1 γ ( 1 s − 1 s + γ )] = 1 γ (1 − e −γt)σ(t), which goes to ∞ when t goes to ∞, and hence is unbounded. This shows that the system is unstable. If γ = 0, the system is called an integrator. Again let u(t) = σ(t). Then, y(t) = L −1 [ 1 s 2 ] = tσ(t), which goes to ∞ as t goes to ∞. This shows that the system is again unstable.
If γ > 0, the impulse response of G(s) is g(t) = e −γtσ(t). Then, |y(t)| = ∫ t 0 g(τ )u(t − τ )dτ ≤ ∫ t 0 |g(τ )||u(t − τ )|dτ. If |u(t)| ≤ M for all t [0, ), then |y(t)| ≤ M ∫ ∞ 0 |g(τ )|dτ = M ∫ ∞ 0 e −γτ dτ = M γ . Hence, as long as u(t) is bounded by M, then y(t) is bounded by M/γ. This shows that the system is stable. In summary, a system with transfer function G(s) = 1 s + γ is stable if γ > 0 and is unstable if γ ≤ 0. Using the procedure in the example to test the stability of a system is often tedious. We desire to have a simple stability test. An LTI system with impulse response function g(t) is stable if and only if ∫ ∞ 0 |g(t)|dt < ∞. (3.1) Proof . Assume that the input u(t) is bounded by M; then |y(t)| = ∫ t 0 g(τ )u(t − τ )dτ ≤ ∫ t 0 |g(τ )||u(t − τ )|dτ ≤ M ∫ ∞ 0 |g(τ )|dτ. Stability and Stabilization If (3.1) is satisfied, choose N = M ∫ ∞ 0 |g(τ )|dτ , and then the input bounded by M generates the output bounded by N; i.e., the system is stable. This shows the sufficiency. If (3.1) is not satisfied, ∫ ∞ 0 |g(t)|dt = ∞. For any M > 0, choose t0 so that ∫ t0 0 |g(τ )|dτ > M and choose u(t) = { 1, if g(t0 − t) ≥ 0; −1, if g(t0 − t) < 0 for 0 ≤ t ≤ t0. Then, |y(t0)| = ∫ t0 0 g(t0 − τ )u(τ )dτ = ∫ t0 0 |g(t0 − τ )|dτ = ∫ t0 0 |g(τ )|dτ > M. This shows that for an input u(t) bounded by 1, the output can be arbitrarily large. Hence, the system is not stable. This proves the necessity. A signal is said to be absolutely integrable over an interval if the integral of the absolute value of the signal over the interval is finite. Hence, a linear system is stable if itsimpulse response is absolutely integrable over [0, ∞). This theorem makes it a bit easier to check the stability of a system.(ii)Defining the functionFand a subset of its domain :pts:F[z_] := (5 - I z)/(5^2 + z^2)pts = {-7, -2, 0, 2, 7};the most straightforward way fulfilling the task is based onParametricPlotandEpilog. Wecan also make a diagram with the basic graphics primitives like e.g. :Line,Circle,Point. Here are the both ways enclosed inGraphicsRow:
GraphicsRow[{ Graphics[{Line[{{0, -0.1}, {0, 0.1}}], Line[{{0, 0}, {0.21, 0}}], Blue, Thick, Circle[{0.1, 0}, 0.1], Red, PointSize[.03], Point[{Re @ #, Im @ #} & /@ F[pts]]}], ParametricPlot[{Re @ #, Im @ #}& @ F[z], {z, -200, 200}, PlotRange -> All, PlotStyle -> Thick, Epilog -> { Red, PointSize[0.03], Point[{Re @ F @ #, Im @ F @ #} & /@ pts]}] }]Studying properties of holomorphic complex mappings is really rewarding, therefore one should take a closer look at it. This function has a simple pole in5 I:Residue[ F[z], {z, 5 I}]-Iand it isconformalin its domain :Reduce[ D[ F[z], z] == 0, z]Falsei.e. it preserves angles locally. One can easily recognize the type ofFevaluatingSimplify[F[z]], namely it is a composition of a translation, rescaling and inversion. We should look at images (viaF) of simple geometric objects. To visualize the structure of the mappingFwechoose an appropriate grid in the complex domain ofFand look at its image. We take a continuous parameter $t$ varying in a range $(-25, 25)$ and contours $\;t+ i\;y $ for $y$ in adiscrete set of values $\{-3, -2,-1, 0, 1, 2, 3 \}$ and another orthogonal contours $\;x+ i\;t$ for $x$ in a discrete set $\{-7,-5,-3, -2, 0, 2, 3, 5, 5\;\}$, i.e.we have a grid of straight lines in the complex plane. Next we'd like to plot the image of this grid through the mapping $F$. Images of every line in the grid will be circles with centers on the abscissa and ordinate respectively intersecting orthogonally. The red points denote values of $F(x)$ on the complex plane for $x$ in $\{-7, -2, 0, 2, 7 \}$. On the lhs we have the original grid in the domain ofFand on the rhs we have the plot of its image :

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