MAE43A Signals & Sysems - Homework, Winer 4 due by he end of class Thursday January 3, 4. Quesion Zener diode malab [Chaparro Quesion.] A zener diode circui is such ha an oupu corresponding o an inpu v s () cos(π) is a clipped sinusoid.5, v s () >.5, x().5, v s () <.5, v s (), else. [Figure. in Chaparro shows he oupu for a few cycles.] Use malab o generae he inpu and oupu signals and plo hem on he same plo for 4 a ime inervals of.. Par (a): Is his sysem linear? Compare he oupu obained from v s () wih ha obained from.3v s (). Par (b): Is he sysem ime-invarian? Explain. Par (c): Is he sysem dynamic (i.e. does i have memory)? Explain. Par (d): Is he sysem sable? Explain. We use malab o generae he inpu and oupu signals and plo hem on he same plo for 4 a ime inervals of.. Here is he malab funcion for Zener diode funcion xzener_diode(v) % a funcion o produc he oupu signal from Zener diode % Here he inpu signal, v, is a scalar and we compue he oupu signal x. xzeros(,); if v>.5 x.5; elseif v<-.5 x -.5; else x v; end reurn Malab code for inpu and oupu signals. % Quesion clear all; clc; [-4:.:4] ; [n,m] size(); x zeros(n,); for i :n x(i,)zener_diode(cos(pi*(i)));
Inpu, Oupu Inpu Oupu.8.6 Inpu and Oupu Signals.4...4.6.8 4 3 3 4 Time (sec) Figure : Inpu and Oupu Signals end plo(,cos(pi*), r ); % inpu curve hold on; plo(,x(:,), LineWidh,); % oupu curve axis([-4 4 -..]);shg; ile( Inpu, Oupu ) xlabel( Time (sec) );ylabel( Inpu and Oupu Signals ) legend( Inpu, Oupu ) grid on Par (a): No, if he sysem S is linear, i should saisfy he propery S[.3v s ()].3S[v s ()]. Check his via malab: clear all; clc; [-4:.:4] ; [n,m] size(); x zeros(n,); y zeros(n,); for i:n x(i) Zener_diode(.3*cos(pi*(i))); % S(.3*Vs()) y(i).3*zener_diode(cos(pi*(i))); %.3*S(Vs()) end subplo(,,); plo(,x); xlabel( Time (sec) );ylabel( S(.3V_s()) ) axis([-4 4 -.4.4]); grid on; subplo(,,); plo(,y); xlabel( Time (sec) );ylabel(.3*s(v_s()) ) axis([-4 4 -.4.4]); grid on; shg; The resul is shown in Figure. The wo signals are clearly differen. Par (b): If he sysem S is ime invarian, i should saisfy S(v( + τ)) x( + τ). We know.5, v s ( + τ) >.5 S(v( + τ)).5, v s ( + τ) <.5 v s ( + τ), else
.4. S(.3V s ())..4 4 3 3 4 Time (sec).4..3*s(v s ())..4 4 3 3 4 Time (sec) Figure : Check lineariy for Zener diode. These figures show ha he sysem is no linear and.5, v s ( + τ) >.5 x( + τ).5, v s ( + τ) <.5 v s ( + τ), else This shows S(v( + τ)) and x( + τ) are exacly he same and he sysem is ime-invarian. Par(c): No. The oupu is an insananeous funcion of he inpu. I does no depend on previous values of he oupu or inpu signals. To see his, consider some fixed ime, T. Consider any wo inpu signals which differ on he ime inerval (, T ) bu are he same on he inerval [T, ). Then he oupus will also be he same on he inerval [T, ). So here canno be any memory in he sysem and i is non-dynamic. Par(d)Yes. This sysem is obviously BIBO sable because a bounded inpu always produces a bounded oupu. Indeed, he oupu is clipped o be bounded in magniude by.5 no maer wheher he inpu is bounded a all. Quesion Analog averaging sysem [Chaparro Quesion.3] Consider he analog averager y() T + T T x(τ) dτ, where x() is he inpu signal and y() is he oupu signal. We consider he signals on he ime domain (, ). Par (a): Find he impulse response, h(), of he averager. Is his sysem causal? Par (b): Le x() (), he uni-sep funcion. Find he oupu, s(), of he averager. [This is he sep response.] Par (c): Verify ha he oupu signal from Par (b) is he inegral of he oupu signal from Par (a). Tha is, s() h(τ) dτ.
impulse response h() T/ T/ sep response y() T/ T/ Figure 3: Quesion-par(a)(b): Impulse response and Sep response Par(a): Impulse response: h() T + T T δ(τ) dτ T, T T, else T [u( + T ) u( T )] The impulse response is shown in Figure 3. Noe ha for he impulse response o be non-zero, he ime needs o belong o he inerval [ T/, + T/]. This gives he limis above. We check he definiion of causaliy: y() does no depend on fuure inpu values. Check: Since y() is relaed o he behavior of x() for [ T, + T ], i depends on fuure inpus. No Saisfied! So, he sysem is no causal. Par(b): Sep response y() T + T T, < T +T/ (τ)dτ T dτ, T T +T/ T T/ dτ, T T, < T T +, T T, T T T ( + T )[u( + T ) u( T )] + u( T ) The sep response is shown in Figure 3. Par(c): If we compare he impulse response and sep response in Figure 3, we are prey sure ha he sep response is he inegral of he impulse response. Verify: Compue s() T [u(τ + T ) u(τ T )]dτ u(τ + T T )dτ T T u(τ T )dτ T τ τ T u(τ + T )dτ + T, T, < T τ τ T u(τ T )dτ T, T, < T () ()
() (), < T s() T ( + T ), T < T ( + T ) ( T ), T T ( + T )[u( + T ) u( T )] + u( T ) This shows ha he oupu signal from Par(b) is he inegral of he oupu signal from Par(a). Quesion 3 Differenial Equaions and Sysems [Robers Quesion 4.4] A well-sirred va has been fed for a long ime by wo sreams of liquid; fresh waer a. cubic meers per second and concenraed blue dye a. cubic meers per second. The va conains cubic meers of his mixure and he mixure is being drained from he va a a rae of.3 cubic meers per second o mainain a consan volume. The blue dye is suddenly replaced by a red dye a he same flow rae. A wha ime afer he swich in dyes does he mixure drawn from he ank have a red-o-blue dye raio of 99:? Le x be he volume of blue dye in he ank. Clearly, x () /3. Bu le us check his laer. We have he following o.d.e. for his volume. dx..3 x, for <, d.3 x, for. Looking a he seady-sae of x ( ), since he sysem has been operaing for a long ime, we have [seady-sae d/d ]..3 x or x /3. For > we have he soluion [o.d.e. plus i.c.] x () x ()e.3 3 e.3. Le x () denoe he volume of red dye. Then x () 3 x (),, for <, [ ] e.3, for. When he mixure drawn have a red-o-blue dye raio of 99:, i saisfies x () 99 x (). Then 3 3 [ e.3 ] 99 3 e.3 53.557 Quesion 4 Time-varying mass of a car [Kamen & Heck Quesion.37] Consider (a badly packed) auomobile wih ime-varying mass M().5 [() ( )].5 ( ).
The model of velociy of he car is M() dv() d +.v() x(), where v() is he velociy of he car, x() is he driving or braking force applied. Compue he velociy v() for all > when v() and x() (), he uni-sep funcion. Your expression for v() should be compleely evaluaed (i.e. all inegrals should be evaluaed). Solve i where M() dv() d +.v() () dv() d M().5, < <.5, +. () v() M() M() Solve his ODE using he inegraing facor mehod for s-order o.d.e.s. The inegraing facor is Define I() exp[ (. M(τ) )dτ] s(τ).. M(τ).5τ, < τ <., τ Calculae inegraing facor remembering ha we are only concerned wih >. I() exp[ s(τ)dτ], [ ]. exp.5τ dτ, < <, [. exp.5τ dτ + ]. dτ,, exp[ ln(.5)], < < ] exp[ ln(.5) +.( )], (.5), < < 4e e., Noe:. dτ. ln(.5τ).5τ.5, [ln(.5) ln ] ln(.5), ( ) τ dτ ln( τ) ln( ) + ln() ln ln(.5).
So, we have v() I() v() (τ) M(τ) I(τ)dτ (.5τ) dτ, < < 3 (.5τ) dτ + 4e e. 3.5 dτ, τ (.5τ), < < (.5τ) + 8e. e., (.5) +, < < 3 4 + 4e., So v() + (.5), < < 4.5(.5) + 4e. (.5),