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<math>H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau</math> | <math>H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau</math> | ||
| − | In this case, we can apply the sifting property to arrive at the system function quite easily. After applying the property, we arrive at <math>H(j\omega)=2-e^{-2j\omega}</math>. One might recognize this is the Laplace transform of the impulse response. | + | In this case, we can apply the sifting property to arrive at the system function quite easily. After applying the property, we arrive at <math>H(j\omega)=2-e^{-2j\omega}</math>. One might recognize this is the Laplace transform of the impulse response evaluated at <math>s=j\omega</math>. |
==Response to a Signal from Question 1== | ==Response to a Signal from Question 1== | ||
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<math>y(t)=(2-e^{6j})(\frac{1+j}{2}e^{-3jt})-(2-e^{4j})(\frac{7}{2j}e^{-2jt})+(2-e^{-4j})(\frac{7}{2j}e^{2jt})+(2-e^{-6j})(\frac{1+j}{2}e^{3jt})</math> | <math>y(t)=(2-e^{6j})(\frac{1+j}{2}e^{-3jt})-(2-e^{4j})(\frac{7}{2j}e^{-2jt})+(2-e^{-4j})(\frac{7}{2j}e^{2jt})+(2-e^{-6j})(\frac{1+j}{2}e^{3jt})</math> | ||
| − | Which a little math becomes | + | Which after a little math becomes |
| − | <math>y(t)=(1+j)e^{-3jt}-\frac{1+j}{2}e^{-3j(t-2)}-\frac{7}{j}e^{-2jt}+\frac{7}{2j}e^{-2j(t-2)}+\frac{7}{j}e^{2jt}-\frac{7}{2j}e^{2j(t-2)}+(1+j)e^{3jt}-\frac{1+j}{2}e^{3j(t-2)}<math> | + | <math>y(t)=(1+j)e^{-3jt}-\frac{1+j}{2}e^{-3j(t-2)}-\frac{7}{j}e^{-2jt}+\frac{7}{2j}e^{-2j(t-2)}+\frac{7}{j}e^{2jt}-\frac{7}{2j}e^{2j(t-2)}+(1+j)e^{3jt}-\frac{1+j}{2}e^{3j(t-2)}</math> |
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| + | It is fairly easy to see that the final <math>y(t)</math> is simply the weighed sum of the doubled input <math>y'(t)=2x(t)</math> and the shifted input <math>y''(t)=x(t-2)</math>. This can serve as confirmation that the system <b>actually was</b> both linear and time invariant. | ||
Latest revision as of 13:40, 26 September 2008
Contents
A Continuous Time, Linear, Time-Invariant System
Consider the system $ y(t)=2x(t)-x(t-2) $.
Unit Impulse Response
Let $ x(t)=\delta(t) $. Then $ h(t)=2\delta(t)-\delta(t-2) $.
System Function
As discussed in class, the system function is
$ H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau $
In this case, we can apply the sifting property to arrive at the system function quite easily. After applying the property, we arrive at $ H(j\omega)=2-e^{-2j\omega} $. One might recognize this is the Laplace transform of the impulse response evaluated at $ s=j\omega $.
Response to a Signal from Question 1
I will use my signal from Question 1.
$ x(t)=7\sin(2t)+(1+j)\cos(3t)=\frac{1+j}{2}e^{-3jt}-\frac{7}{2j}e^{-2jt}+\frac{7}{2j}e^{2jt}+\frac{1+j}{2}e^{3jt} $
Since we have represented the original signal as a sum of complex exponentials, we simply have to multiply each term of the original input signal by $ H(j\omega) $ to obtain the corresponding term in the output.
$ y(t)=H(jw)x(t) $
$ y(t)=(2-e^{6j})(\frac{1+j}{2}e^{-3jt})-(2-e^{4j})(\frac{7}{2j}e^{-2jt})+(2-e^{-4j})(\frac{7}{2j}e^{2jt})+(2-e^{-6j})(\frac{1+j}{2}e^{3jt}) $
Which after a little math becomes
$ y(t)=(1+j)e^{-3jt}-\frac{1+j}{2}e^{-3j(t-2)}-\frac{7}{j}e^{-2jt}+\frac{7}{2j}e^{-2j(t-2)}+\frac{7}{j}e^{2jt}-\frac{7}{2j}e^{2j(t-2)}+(1+j)e^{3jt}-\frac{1+j}{2}e^{3j(t-2)} $
It is fairly easy to see that the final $ y(t) $ is simply the weighed sum of the doubled input $ y'(t)=2x(t) $ and the shifted input $ y''(t)=x(t-2) $. This can serve as confirmation that the system actually was both linear and time invariant.
