Find the amplitude for a system.

[Aedrha2]

Hi there, I'm studying a course in signal analysis and have come across an exercise where I find the math a bit tricky.

I am to find the amplitudefunction |H( f )| for a system with the impulseresponse [math]h(n)=\delta(n)+0.9\delta(n-D)[/math]Where D= 500

I did some transforms: [math]h(n)=\delta(n)+0.9\delta(n-D)\\ H(z)=1+0.9z^{-D}\\ z=e^{j\omega}\\ H(\omega)=1+0.9e^{-j\omega D}[/math]
My thought was to factor out [math]e^{-jxD\omega}[/math] for some factor x and then use eulers formula. But the 0.9 term is holding me back!

I'd greatly appreciate any help or tips!

 

[Deleted member 4993]

Hi there, I'm studying a course in signal analysis and have come across an exercise where I find the math a bit tricky.

I am to find the amplitudefunction |H( f )| for a system with the impulseresponse [math]h(n)=\delta(n)+0.9\delta(n-D)[/math]Where D= 500

I did some transforms: [math]h(n)=\delta(n)+0.9\delta(n-D)\\ H(z)=1+0.9z^{-D}\\ z=e^{j\omega}\\ H(\omega)=1+0.9e^{-j\omega D}[/math]
My thought was to factor out [math]e^{-jxD\omega}[/math] for some factor x and then use eulers formula. But the 0.9 term is holding me back!

I'd greatly appreciate any help or tips!

Have you considered:

0.9 = e^ln(0.9)

 

[Aedrha2]

Have you considered:

0.9 = e^ln(0.9)

I tried as recommended 0.9 = e^ln(0.9)

[math]1+0.9e^{-j\omega D}\\ 1+e^{ln0.9}e^{-j\omega D}\\ 1+e^{ln0.9-j\omega D}\\ e^{-\frac{1}{2}(ln0.9-j\omega D)}(e^{\frac{1}{2}(ln0.9-j\omega D)}+e^{-\frac{1}{2}(ln0.9-j\omega D)})[/math]
This gets me closer to what i was getting at but nit quite there.

My thought was to get something like [imath]H(\omega)=e^{\omega j}(1+\cos(\omega))[/imath]. Now it seems to me that the ln0.9 term is in the way.

 

[Deleted member 4993]

H(ω)=1+0.9e^−jωD

1. Write the above in a + jb form
   
   
2. Write A(mplitude) = √(a² + b²)