I'm not sure how they got the answer (provided below) for the following discrete system

y[n] = x[n]- 2x[n-2]+ x[n-3]- 3x[n-4]

the answer is given as .... h[n] = [1 0 -2 1 -3]

why is this? any help would be greatly appreciated.

Thank you.

  • 2
    \$\begingroup\$ They fed an unit impulse to the system and the output response for an impulse is a vector with the FIR coefficients of the system. \$\endgroup\$
    – Justme
    Nov 19 '18 at 19:46
  • \$\begingroup\$ Where abouts did the zero come from? thanks \$\endgroup\$
    – wwjrmmr
    Nov 19 '18 at 20:03
  • 3
    \$\begingroup\$ Because there is 0*x[n-1] in the system. \$\endgroup\$
    – Justme
    Nov 19 '18 at 20:10

When the difference equation is given, one can directly read the impulse response from it.

In your case the difference equation is given as: y[n] = x[n]- 2x[n-2]+ x[n-3]- 3x[n-4]

This means the output is the input + minus twice the input delayed by two samples + the input delayed by three samples... Notice, the input delayed one sample has no influence on the output.

When stimulating the input of this LTI system with a unit impulse y[0] = x[0]. All other terms are zero since our input is an impulse which is a dirac at 1 and 0 else. At timepoint 1 y[1] becomes 0, since there is no x[n-1] term. Try to put in 1 for every n. Remember x is zero everywhere except at 0. At timepoint 2 y[2] is -2x[2-2]=-2x[0]=2. Time steps y[3] becomes x[3-3]=x[0]=1 and And for the last term after four steps -3x[4-4]=-3x[0]=-3. After that y is zero.

Conclusion: stimulating the network with a impulse the response becomes h[n] = [1 0 -2 1 -3]. One gets this result quite fast if he takes all coefficients of the delayed x terms.


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