Impulse response of an accumulator

[jagsee1972]

I am a bit stuck understanding why the impulse response of an accumulator function is the unit step function: u[n].

The impulse response h[n] of a digital system can be obtained by applying the unit impulse function δ[n].

If the system is an accumulator that is in the form: \[\sum_{k = -\infty}^\infty [n]\] it was supposed to be x[n]. if now x[n] is a discrete function which has an ampitude of 2 when n=0, then its impulse response would be 2*u[n]. Its always written as just u[n] is it assumed that it may need to be multiplied by some constant if x[n] is not 1 and n=0.

I would really a[preciate some advice on this matter, thanks in advance.

 

[axcdd]

unit impulse function δ[n] is 1 when n=0 and 0 to n !=0.
its called Kronecker delta (https://en.wikipedia.org/wiki/Kronecker_delta)

for FIR filters generally y[n] = x[n] * h[n]; when you take δ[n] as an input (x[n]=δ[n]) u got y[n] = δ[n]*h[n]. then δ[n] = 1 when n=0 and 0 when n /=0 so y[n]=h[n]

 

[jagsee1972]

Thanks for the reply axcdd, however I am still having some trouble understanding.

for example if x[n] included the following aseries of values: x[n] = {3, 4, 6, 7, 8} where amplitude 6 is when n=0;

The digital system is an accumulator \[DS=\sum_{n = \infty}^\infty [n]\]

Sorry its supposed to be x[n], can not get to grips with latex code.

Then the impulse response of this function can be obtained by applying a unit impulse function to the Digital System (DS). Is this not the same as multiplying the accumulator function by the unit impulse function. If so this means that all samples other than 0, will equal 0. However the 0th sample will equal \[ 6\delta[n]\] and h[n] will equal 6u[n].

THis is the result I get, however in literature the impulse response is u[n], is it just assumed that it may need multiplying by a constant or have I misunderstood something.

Jag.

 

[axcdd]

yes your right because u[n] = 0 for n < 0; 1 for n >=0.
so if you use 6*δ[n] then DS = 6*u[n] for n>=0; ( But as you Input 6*δ[n] your system response is still u[n] thou output is equall to 6*u[n])

In books there is always definition that one should measure Linear Time-Invariant System by applying δ[n].

some simple lecture.
https://dsp7.ee.uct.ac.za/~nicolls/lectures/eee401f/02_dts_2up.pdf

 

[jagsee1972]

Thanks axcdd. I did not know there was a difference between system response and output. Do now though, thanks a lot