unit impulse dimension

Thread Starter

screen1988

Joined Mar 7, 2013
310
What does this mean?
vIN is a unit impulse at time t = 0.
vIN(t)=1δ(t) volt-seconds
I don't understand the dimension here. vIN is a voltage, it has to have the dimension of Volt (V).
1δ(t) has to have the dimension of volt?
If so, what is the dimension of 1 and δ(t)?
 

t_n_k

Joined Mar 6, 2009
5,455
The area under the impulse voltage vs time curve equates to units of volt-seconds.

For example a rectangular pulse of height 1 volt and duration of 1 second has an area under the curve of 1 volt-second.

A unit impulse presumably also has an area under the curve of 1 volt-second. This is hard to visualize since the [unit] impulse has infinitely small time duration. However this notion provides a means of "physically" quantifying the impulse function.
 

Thread Starter

screen1988

Joined Mar 7, 2013
310
But volt-seconds is not the unit of voltage.
vIN(t)=1δ(t) volt-seconds
I understand you said about unit impulse but how the formula above correct.
The right hand side has the unit of volt-seconds.
The left hand side has the unit of volt.
 

t_n_k

Joined Mar 6, 2009
5,455
But volt-seconds is not the unit of voltage.
vIN(t)=1δ(t) volt-seconds
I understand you said about unit impulse but how the formula above correct.
The right hand side has the unit of volt-seconds.
The left hand side has the unit of volt.
Other than what I have already stated I can't comment on the specific information you originally posted, as I am not acquainted with the original source of the material or its context.

It's worth considering the broader scope of the Dirac function which isn't quite as simple a concept as you might think. Even the term function in relation to Dirac function doesn't make sense in the strictest interpretation of the meaning ...

See http://en.wikipedia.org/wiki/Dirac_delta_function
 
Last edited:

WBahn

Joined Mar 31, 2012
27,893
The answer depends on what units you ascribe to the delta function itself. There are two possibilities that make sense, either that the delta function itself is dimensionless, or that it's integral over the entire domain of the function is dimensionless -- since mathematicians pretty much only deal with pure numbers, there is no clear guidance from a math perspective.

If your domain variable is not dimensionless, then you have to choose. If your function is in the time domain, then you have the choice of declaring that the delta function has units of inverse time and its integral has one unit of dimensionless area, or you can declare that the delta function is dimensionless and that its integral has one unit of time. Both definitions are used and as long as you are self-consistent you won't run into problems. But once you choose, you have to make all of your other coefficients consistent with that choice.

Thus, in your case you can have either:

vIN(t)=1V δ(t)

or

vIN(t)=1Vs δ(t)

In the first case, the delta function is dimensionless and in the second it has units of inverse-seconds. In either case, the units of vIN are volts.

One of the things to keep in mind is that as soon as you are talking about a function that has a delta function in it, you aren't talking about a real function, but rather a theoretical distribution and the only thing that has any real meaning is the integral of the function.
 

t_n_k

Joined Mar 6, 2009
5,455
Sometimes an example can be informative.

Consider the case of a previously unenergized series RL circuit excited by a unit delta function.

At the instant the delta impulse is active at t=0- we may regard the inductor as an open circuit supporting the entire impulse voltage. A current will be established in the inductor over the active impulse interval t=0- to t=0+. The current established at time t=0+ will be

\(i_{L(0^+)}=\frac{1}{L}\int_{0^-}^{0^+} \delta(t) dt\)

The term involving integral of the unit delta function reduces to unity - being the summation of the area under the unit delta function.

So we may conclude that the current in the inductor at t=0+ will simply be

\(i_{L(0^+)}=\frac{1}{L}\)

This initial current will eventually "discharge" through the series resistance R according to the relationship.

\(i_L(t)=\frac{1}{L}e^{-\frac{Rt}{L}} \ \text{for t > 0}\)
 
Last edited:

WBahn

Joined Mar 31, 2012
27,893
Sometimes an example can be informative.

Consider the case of a previously unenergized series RL circuit excited by a unit delta function.

At the instant the delta impulse is active at t=0- we may regard the inductor as an open circuit supporting the entire impulse voltage. A current will be established in the inductor over the active impulse interval t=0- to t=0+. The current established at time t=0+ will be

\(i_{L(0^+)}=\frac{1}{L}\int_{0^-}^{0^+} \delta(t) dt\)
But you KNOW that this is wrong because the units don't work out. The term 1/L does NOT have units of current (it has units of Vs/A).
 

Thread Starter

screen1988

Joined Mar 7, 2013
310
The term 1/L does NOT have units of current (it has units of Vs/A).
I think 1/L has unit 1/Ω.s and \( i_{L}\) has unit of A or V/Ω
=> \(\int _{0-}^{0^{+}}\delta dt\) has unit of V.s and it seems to fit with the unit (volt-seconds) chose in my first post.
 

t_n_k

Joined Mar 6, 2009
5,455
Yep - I would imagine with δ(t) being an infinite voltage impulse the integral term would reduce to units of Volt-sec which would satisfy the units matter.
 

WBahn

Joined Mar 31, 2012
27,893
Yep - I would imaginewith δ(t) being an infinite voltage impulse the integral term would reduce to units of Volt-sec which would satisfy the units matter.
And so now

\(i_{L(0^+)}=\frac{1}{L}\)
is somehow dimensionally consistent?

That's like saying that a pan is two gallons deep or that my house is tweny coulombs high. What does it mean for a current to be one over two henries?

The cleanest way to handle things is to not get sucked into giving a delta function units. There is no need to. Otherwise, you have to keep track that this delta function has these units that delta function has those units. We don't do that with transcendental functions, such as sine or log, so why do it with the delta? Instead, just like the transcendental functions, use coefficients to carry those units.

If you want a voltage impulse that, integrated across the origin, gives an inductor an initial current corresponding to an impulse of 1Vs, then you apply a delta function that is scaled by the size of the impulse per unit od the domain variable. In this case, you want an impulse of 1Vs and your domain variable is seconds, so your scaling factor is 1V. Hence

\(
i_{0+}\;=\;i_{0-}\;+\;\frac{1}{L} \int_{0-}^{0+}1V \delta (t) dt
\)

\(
i_{0+}\;=\;i_{0-}\;+\;\frac{1V}{L} \int_{0-}^{0+} \delta (t) dt
\)

Since δ(t) is dimensionless, the integral over any time that includes the origin is 1s, hence

\(
i_{0+}\;=\;i_{0-}\;+\;\frac{1Vs}{L}
\)

If, as stated, the inductor is initially unenergized, then the current just prior to the application of the impulse is zero and we are left with

\(
i_{0+}\;=\;\frac{1Vs}{L}
\)

Impulses have units and just because the magnitude happens to be unity does not mean that we can throw away the units. A desk that is 1 yard tall can't just be called a unit high desk.
 
Top