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What will be the value of this signal (specified as impulse train) , say for values of t from 0 to 6 $$g(t)=\sum_{k=- \infty }^\infty \delta(t-2k)$$

mahes
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1 Answers1

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The Dirac impulse $\delta(t)$ is a distribution and not a function in the usual sense. Its value at $t=0$ is not defined. However, $\delta(t)=0$ for $t\neq 0$. In your case, you have an impulse train, i.e. a sequence of impulses at $t=2k$, i.e. at points where $t$ is an even integer. So the impulse train has value $0$ almost everywhere except at points $t=2k$, where its value is not defined. Usually an impulse train is "plotted" like this:

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In your case the distance between the impulses is $T=2$.

EDIT:

Judging from your comment you need the derivative of a signal which jumps up by a value of $3$ if $t$ is an even integer (i.e., $t=2k$) and which jumps down by $3$ if $t$ is an odd integer (i.e., $t=2k+1$). In this case the derivative is given by two impulse trains, on with impulses of area $3$ at $t=2k$, and one with impulse of area $-3$ at $t=2k+1$, i.e.

$$g(t)=3\sum_{k=-\infty}^{\infty}\delta(t-2k)-3\sum_{k=-\infty}^{\infty}\delta(t-2k-1)$$

Matt L.
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  • says the signal g(t) is derivative of x(t). where $x(t)=1$; for $0 \leq t \leq 1$ and $x(t)=-2$; for $1<t<2$. my problem is based on a question from ALAN V OPPENHEIM's Signals and System book. – mahes Jun 11 '14 at 09:55
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    @user2332665: The signal $g(t)$ as given in your question is the derivative of a signal which increases by 1 at every even integer $t$. – Matt L. Jun 11 '14 at 10:04
  • @user2332665: I added some info to my answer concerning the derivative of the signal you described in your comment. – Matt L. Jun 11 '14 at 10:09
  • would you mind if I show you the full problem. Problem – mahes Jun 11 '14 at 14:50
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    @user2332665: The answer is given in my edited answer above. At $t=2k$ the signal needs to go up to $+1$ (from $-2$), and at odd integers $t=2k+1$ it needs to go down to $-2$. Since the difference is always $3$, the impulses need to be multiplied by $3$ (going up by $+3$, going down by $-3$). So $A_1=3$, $A_2=-3$, $t_1=0$, $t_2=-1$. – Matt L. Jun 11 '14 at 15:06