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UIUC ECE 210 L47-2

UIUC ECE 210 L47-2

... to the minus st you get 0 minus

UIUC ECE 210 L29-2

UIUC ECE 210 L29-2

1 over minus j n

UIUC ECE 210 L49-2

UIUC ECE 210 L49-2

So only those

UIUC ECE 210 L46-2

UIUC ECE 210 L46-2

This Fourier transform to

UIUC ECE 210 L51-2

UIUC ECE 210 L51-2

UIUC ECE 210 L51-2

UIUC ECE 210 25-2

UIUC ECE 210 25-2

... to here I have

UIUC ECE 210 24-2

UIUC ECE 210 24-2

Minus

UIUC ECE 210 L47-1

UIUC ECE 210 L47-1

Okay so this will retransform to that and then you're given that the input ft is a sine of

UIUC ECE 210 5-2

UIUC ECE 210 5-2

I1 times

UIUC ECE 210 L27-2

UIUC ECE 210 L27-2

He is

ECE 210: Analog Signal Processing

ECE 210: Analog Signal Processing

ECE 210

UIUC ECE 210 2-2

UIUC ECE 210 2-2

No okay you solve these

UIUC ECE 210 7-2

UIUC ECE 210 7-2

And i don't know exactly if the