Convolution and Transfer Function

[roam]

Homework Statement

The impulse response $h(t)$ of a linear time invariant system is real valued. Where * denotes the complex conjugate, the transfer function satisfies:

$$H(- \nu) = H^* (- \nu)$$

Use this result to show that for such a system, given the input $f(t)= \sin(2 \pi \nu t)$, the output will be:

$$g(t)=(f*h)(t) = |H(\nu)| \sin (2\pi \nu t +arg(H(\nu)))$$

Homework Equations

Convolution integral is given by: $g(t)= \int^\infty_{- \infty} f (\tau) h(t-\tau) d \tau$

For input $f(t)=\exp (j2 \pi \nu t)$:

$g(t)=(f*h)(t)= (\int h(\tau) \exp (- j 2 \pi \nu \tau) d \tau) \ exp (j 2 \pi \nu t) = H (\nu) \exp (j 2 \pi \nu t)$

The Attempt at a Solution

So, starting from the definition of a convolution:

$$\int^{\infty}_{-\infty} \sin (2\pi \nu \tau) . h(t- \tau) d \tau \ \ (1)$$

We must show that this is equal to:

$$g(t)= |H(\nu)| \sin (2\pi \nu t +arg(H(\nu)))$$

I know that $H(\nu)= \int^{\infty}_{-\infty} h(t) e^{-2 \pi j \nu t} dt$, so the expression above is equal to:

$$g(t)= (\int^{\infty}_{-\infty} h(t-\tau) e^{-2 \pi j \nu \tau} d \tau) \sin (2\pi \nu t +arg(H(\nu)))$$

So, how can I manipulate (1) further to get to this result?

Any explanation is greatly appreciated.

 

[blue_leaf77]

$H(- \nu) = H^* (- \nu)$

Are you sure the minus signs should appear in the argument in both sides of the equation? If yes, that tells you that $H(\nu)$ is real, but then the placement of the minus signs is superfluous.
Further consequence is that $\textrm{arg }H(\nu)=n\pi$ with $n$ integer numbers.

 

[roam]

Are you sure the minus signs should appear in the argument in both sides of the equation? If yes, that tells you that $H(\nu)$ is real, but then the placement of the minus signs is superfluous.
Further consequence is that $\textrm{arg }H(\nu)=n\pi$ with $n$ integer numbers.

Oops, that was a typo, sorry. I meant to say:

$$H(- \nu) = H^* (\nu)$$

How do we use this to get to the given expression for $g(t)$?

 

[blue_leaf77]

Start by expressing $\sin(2\pi \nu t)$ in terms of complex exponentials and $h(t-\tau)$ in terms of its Fourier transform, then plug into equation (1).

 

[roam]

Start by expressing $\sin(2\pi \nu t)$ in terms of complex exponentials and $h(t-\tau)$ in terms of its Fourier transform, then plug into equation (1).

What do you mean by expressing $h(t-\tau)$ in terms of its Fourier transform? In terms of complex exponentials $\sin(2\pi \nu t) = \frac{e^{j 2\pi \nu t} - e^{-j 2\pi \nu t}}{2 j}$. So plugging this into Eq (1):

$$g(t) = \int^{\infty}_{-\infty} \frac{e^{j 2\pi \nu \tau} - e^{-j 2\pi \nu \tau}}{2 j} . h(t-\tau) \ d \tau$$

$$= \frac{-j}{2} \left( \underbrace{\int^{\infty}_{-\infty} h (t-\tau) e^{+ j 2 \pi \nu \tau} \ d \tau}_{\text{inverse transform}} - \underbrace{\int^{\infty}_{-\infty} h (t-\tau) e^{-j 2 \pi \nu \tau} \ d \tau}_{\text{Fourier transform of h}}\right)$$

We can rewrite this in terms of the transfer function as:

$$= \frac{-j}{2} \left( \int^{\infty}_{-\infty} h (t-\tau) e^{+ j 2 \pi \nu \tau} \ d \tau - H(\nu) \right)$$

Is that right? I am stuck here, I don't know how to progress from here.

 

[blue_leaf77]

$$= \frac{-j}{2} \left( \underbrace{\int^{\infty}_{-\infty} h (t-\tau) e^{+ j 2 \pi \nu \tau} \ d \tau}_{\text{inverse transform}} - \underbrace{\int^{\infty}_{-\infty} h (t-\tau) e^{-j 2 \pi \nu \tau} \ d \tau}_{\text{Fourier transform of h}}\right)$$

No, that's not exactly the Fourier transform of $h(t)$, instead they are the FT of the shifted $h(t)$, and obviously this is different from $H(\nu)$.

What do you mean by expressing h(t−τ)h(t−τ)h(t-\tau) in terms of its Fourier transform?

I mean the inverse of $H(\nu)= \int^{\infty}_{-\infty} h(t) e^{-2 \pi j \nu t} dt$.

 

[roam]

According to the shifting property of the Fourier transform: $f(t-t_0) \iff exp(-j2 \pi \nu t_0) F(\nu)$, the FT of h(t-τ) becomes:

$$H(t- \tau) = \exp(-j 2 \pi \nu \tau) \left( \int^{\infty}_{- \infty} h(t) \exp(-j 2 \pi \nu t) d t \right)$$

Substituting into (1) we find the equation:

$$g(t) = \int^{\infty}_{- \infty} \frac{e^{j 2 \pi \nu t} - e^{-j 2 \pi \nu t}}{2 j} . \left( \exp(-j 2 \pi \nu \tau) \left( \int^{\infty}_{- \infty} h(t) \exp(-j 2 \pi \nu t) d t \right) \right) dt$$

The last term, $\int^{\infty}_{- \infty} h(t) \exp(-j 2 \pi \nu t) d t$, is equal to $H(t)$. But what can I do with the other terms?

How can we get to $\sin(2 \pi \nu t + arg(H(\nu)))$?

I mean the inverse of $H(\nu)= \int^{\infty}_{-\infty} h(t) e^{-2 \pi j \nu t} dt$.

The inverse of $H(\nu)$ is the complex conjugate given by:

$$H^* (\nu)= H(- \nu) = \int^{\infty}_{- \infty} h(-t) e^{-2 \pi j \nu t} dt$$

I'm not sure how we can introduce this into the equation for convolution.

 

[Charles Link]

Perhaps I can give a helpful input here. In F.T. theory, when $g(t)=\int h(t-t')f(t')dt'$ in F.T. space the result is $g(\omega)=h(\omega)f(\omega)$ Your notation is slightly different=you're using $H(\nu)$ instead of $h(\omega)$ but I think you should be able to follow this part. When $f(t)$ is a sinusoid, $f(\omega)$ becomes a delta function expression. (Comes from $\delta(\omega)=(1/2\pi)\int exp(i \omega t) d t$ ) Once you get $g(\omega)$ you need to take the inverse transform to get the expression for g(t)

 

[blue_leaf77]

$$H(t- \tau) = \exp(-j 2 \pi \nu \tau) \left( \int^{\infty}_{- \infty} h(t) \exp(-j 2 \pi \nu t) d t \right)$$

Substituting into (1) we find the equation:

$$g(t) = \int^{\infty}_{- \infty} \frac{e^{j 2 \pi \nu t} - e^{-j 2 \pi \nu t}}{2 j} . \left( \exp(-j 2 \pi \nu \tau) \left( \int^{\infty}_{- \infty} h(t) \exp(-j 2 \pi \nu t) d t \right) \right) dt$$

No, you have got it wrong there, You were on the right track when introducing the use of shifting theorem, but you used it in a wrong way.
Start from the inverse transform
$$
h(t)= \int^{\infty}_{-\infty} H(\nu) e^{2 \pi j \nu t} d\nu
$$
Since you what you want is $h(t-\tau)$, you have got to replace $t$ in the above equation with $t-\tau$. Having done this, plug into equation (1).

The last term, $\int^{\infty}_{- \infty} h(t) \exp(-j 2 \pi \nu t) d t$, is equal to $H(t)$.

No, it's not equal to $H(t)$, it's equal to $H(\nu)$.

EDIT: There is a mistake in the inverse integral transform above - the integral element $dt$ has been replaced by $d\nu$, which is the correct one.

 

[roam]

Start from the inverse transform
$$
h(t)= \int^{\infty}_{-\infty} H(\nu) e^{2 \pi j \nu t} dt
$$
Since you what you want is $h(t-\tau)$, you have got to replace $t$ in the above equation with $t-\tau$. Having done this, plug into equation (1).

Thank you very much for the hint. But I'm not sure what is wrong here because when I plug this into (1):

$$g(t)= \frac{-j}{2} \int^{\infty}_{-\infty} \left( e^{j2 \pi \nu t} - e^{-j2 \pi \nu t} \right) \left( \int^{\infty}_{-\infty} H(\nu) e^{2 \pi j \nu (t - \tau)} d(t-\tau) \right) d \tau,$$

The expression for the inverse transform goes to infinity when I integrate over $(-\infty, \infty)$, because we will have something like $\frac{H(\nu)}{e^{2 \pi j \nu}} [e^{\infty} - e^{-\infty}]$.

And also where can we use the H(-ν)=H^*(ν) property?

 

[blue_leaf77]

I edited my post #9, there is a small mistake there.

Thank you very much for the hint. But I'm not sure what is wrong here because when I plug this into (1):
$$g(t)= \frac{-j}{2} \int^{\infty}_{-\infty} \left( e^{j2 \pi \nu t} - e^{-j2 \pi \nu t} \right) \left( \int^{\infty}_{-\infty} H(\nu) e^{2 \pi j \nu (t - \tau)} d\nu \right) d \tau,$$

There is still a mistake originating from your side in the above equation, remember that you should be working with $\sin(2\pi \nu \tau)$, not $\sin(2\pi \nu t)$. (I have purposely corrected another mistake which is due to me, namely the integral element from $d(t-\tau)$ to $d\nu$ for you).

The next step is to reverse the integration order, where you will integrate over $d\tau$ first. In order to do this, collect all terms which contains $\tau$ in one.

And also where can we use the H(-ν)=H*(ν) property?

That comes into play later.

 

[Charles Link]

The final result of the calculation is a very useful result in linear response theory, that the response g(t) of any linear system to a sinusoidal input f(t) is a sinusoidal g(t) at the same frequency that has an amplitude factor and phase change that can be found from the F.T. of the response function h(t). This result has many applications including AC circuit theory where the output voltage of a sinusoidal input voltage is a sinusoidal output voltage with the various impedance factors making up the $H(\nu)$. A couple of algebraic errors made your calculations difficult to follow (e.g. in post #7 $H(t-\tau)$ should read $H(\nu,\tau)$) , but suggest you use the convolution theorem result for Fourier transforms mentioned in post #8. This one should only need a couple of lines of algebra and the final result should emerge.

 

[roam]

I edited my post #9, there is a small mistake there.

There is still a mistake originating from your side in the above equation, remember that you should be working with $\sin(2\pi \nu \tau)$, not $\sin(2\pi \nu t)$. (I have purposely corrected another mistake which is due to me, namely the integral element from $d(t-\tau)$ to $d\nu$ for you).

The next step is to reverse the integration order, where you will integrate over $d\tau$ first. In order to do this, collect all terms which contains $\tau$ in one.

What do you mean by collecting τ terms in one?

So we had:

$$g(t) = \frac{1}{2j} \int^{\infty}_{-\infty} (e^{j 2 \pi \nu \tau} + e^{-j 2 \pi \nu \tau}) \left( \int^{\infty}_{-\infty} H(\nu) e^{2 \pi j \nu (t-\tau)} d \nu \right) d \tau$$

Do you mean that I should bring H(ν) out of the integral and then integrate $e^{2 \pi j \nu (t-\tau)}$ over $d\tau$?

That comes into play later.

We were told that we can use a property like this: $A e^{j 2 \pi \nu t} + B e^{-j 2 \pi \nu t} \to A H(\nu) e^{j 2 \pi \nu t} + B \underbrace{H(- \nu) e^{-j 2 \pi \nu t}}_{\text{H*(ν)}}$. Is that right?

I'm not sure how to justify this result. But if we can use that, then we can introduce $H^*(\nu)$ into the first part of the equation like this:

$$\frac{1}{2j} \left( H(\nu) e^{j 2 \pi \nu t} - H^* (\nu) e^{-j 2 \pi \nu t} \right)$$

If this the correct idea?

 

[blue_leaf77]

Before proceeding, I forgot to warn you to use different notation between the frequency of $f(t)$ and the frequency components of $h(t-\tau)$. With this taken into account, the equation becomes
$$
g(t) = \frac{1}{2j} \int^{\infty}_{-\infty} (e^{j 2 \pi \nu \tau} - e^{-j 2 \pi \nu \tau}) \left( \int^{\infty}_{-\infty} H(\nu') e^{2 \pi j \nu' (t-\tau)} d \nu' \right) d \tau
$$

$$g(t) = \frac{1}{2j} \int^{\infty}_{-\infty} (e^{j 2 \pi \nu \tau} + e^{-j 2 \pi \nu \tau}) \left( \int^{\infty}_{-\infty} H(\nu) e^{2 \pi j \nu (t-\tau)} d \nu \right) d \tau$$

Can you please stop making typos?

Do you mean that I should bring H(ν) out of the integral and then integrate e2πjν(t−τ)e2πjν(t−τ)e^{2 \pi j \nu (t-\tau)} over dτdτd\tau?

For example, from the above integral you will get two terms, the first one being $\int\int e^{j2\pi\nu \tau}H(\nu')e^{2\pi j \nu'(t-\tau)}d\nu' d\tau$. Rearrange the order to obtain
$$
\int H(\nu') e^{2\pi j \nu' t}\left( \int e^{j2\pi \tau(\nu-\nu')} d\tau \right) d\nu'
$$
Then solve the integral inside the bracket using the definition of a delta function.

We were told that we can use a property like this: $A e^{j 2 \pi \nu t} + B e^{-j 2 \pi \nu t} \to A H(\nu) e^{j 2 \pi \nu t} + B \underbrace{H(- \nu) e^{-j 2 \pi \nu t}}_{\text{H*(ν)}}$. Is that right?

I'm not sure how to justify this result. But if we can use that, then we can introduce $H^*(\nu)$ into the first part of the equation like this:

$$\frac{1}{2j} \left( H(\nu) e^{j 2 \pi \nu t} - H^* (\nu) e^{-j 2 \pi \nu t} \right)$$

If this the correct idea?

Yes, that's correct. But you don't instantly arrive at that equation, do you?

 

[roam]

Rearrange the order to obtain
$$
\int H(\nu') e^{2\pi j \nu' t}\left( \int e^{j2\pi \tau(\nu-\nu')} d\tau \right) d\nu'
$$
Then solve the integral inside the bracket using the definition of a delta function.

Thank you, I understand now. The entire expression for g(t) would then be:

$$\frac{1}{2j} \left( \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t}\left( \int^{\infty}_{-\infty} e^{j2\pi \tau(\nu-\nu')} d\tau \right) d\nu' \right) - \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t} \left( \int^{\infty}_{-\infty} e^{- j2\pi \tau(\nu-\nu')} d\tau \right) d\nu' \right) \right)$$

Could you please explain how the delta function factors into the equation?

Do you mean the Dirac delta function $\delta(t)= \left\{\begin{matrix}\infty, \ \ t=0 \\ 0, \ \ else \end{matrix}\right.$?

 

[blue_leaf77]

Could you please explain how the delta function factors into the equation?

Take a look at Charles's post #8.

 

[roam]

Take a look at Charles's post #8.

I can't follow Charles's post. He says that if f(t) is sinusoidal f(ω) becomes a delta function expression, so that means you would rewrite g(ω)=h(ω)f(ω) as h(ω)δ(ω)?

A little bit more explanation would be appreciated because I am not sure how exactly it's applicable to this problem (solving the integral $\int^{\infty}_{-\infty} e^{j2 \pi \tau (\nu-\nu')}$).

 

[blue_leaf77]

It's this

$\delta(\omega)=(1/2\pi)\int exp(i \omega t) d t$

Probably you can neglect the constant prefactor for the present purpose.

$$\frac{1}{2j} \left( \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t}\left( \int^{\infty}_{-\infty} e^{j2\pi \tau(\nu-\nu')} d\tau \right) d\nu' \right) - \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t} \left( \int^{\infty}_{-\infty} e^{- j2\pi \tau(\nu-\nu')} d\tau \right) d\nu' \right) \right)$$

Seems like your sloppy nature in manipulating mathematical expression cannot disappear anytime soon. Check again the second term.

 

[Charles Link]

I can't follow Charles's post. He says that if f(t) is sinusoidal f(ω) becomes a delta function expression, so that means you would rewrite g(ω)=h(ω)f(ω) as h(ω)δ(ω)?

A little bit more explanation would be appreciated because I am not sure how exactly it's applicable to this problem (solving the integral $\int^{\infty}_{-\infty} e^{j2 \pi \tau (\nu-\nu')}$).

By delta function, I meant an expression containing delta functions. In this case I think you're going to get a $\delta(\nu-\nu_o)$ plus a $\delta(\nu+\nu_o)$ . In the original sinusoid in time you need to call the frequency $\nu_o$ to indicate it is a constant. You sort of got around this by introducing $\nu'$ but the prime is a little clumsy. You are better off from the beginning to call the constant sinusoidal input frequency $\omega_o$ or $\nu_o$. When you have a $\delta(\nu)$ instead of $\delta(\omega)$ where $\omega=2 \pi \nu$ I believe blue_leaf77 is correct, the $1/(2 \pi)$ factor will be absent from the delta function formula. Suggestion is to work the Fourier transform of $\sin(2 \pi \nu_o t)$ and then I think you will find it all comes together. (Express the sine function in terms of complex exponentials and then take the F.T.)

 

[roam]

There was a sign error, this should be correct:

$$g(t)=\frac{1}{2j} \left( \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t}\left( \int^{\infty}_{-\infty} e^{j2\pi \tau(\nu-\nu')} d\tau \right) d\nu' \right) - \left( \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t} \left( \int^{\infty}_{-\infty} e^{- j2\pi \tau(\nu+\nu')} d\tau \right) d\nu' \right) \right)$$

According to my post #13 the equation has to be reduced to the form $\frac{1}{2j} \big[H(\nu)e^{+...}-H(\nu) e^{-...} \big]$.

I wasn't sure how to use the delta property but I found these properties in a textbook:

$$\int^{\infty}_{-\infty} e^{-j 2 \pi (\nu-\nu_0)t} dt = \delta (\nu - \nu_0) \ \ (i)$$

$$FT(\delta(t-t_0)) = e^{-j2 \pi \nu_0 t_0} \ \ (ii)$$

For instance using (i), I think the second term in the equation becomes:

$$- \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t} . \delta(\nu + \nu') d \nu'$$

This now looks similar to the inverse FT of $\delta(\nu + \nu')$. If we had a $e^{-2\pi j \nu' t}$ instead we could use property (ii) and write:

$$=- H(\nu) e^{-j 2 \pi \nu' t}$$

So what do I need to do in this case? This is very confusing and unfortunately the textbook does not explain this.

Likewise, when we deal with the first term in the equation, we can't use (i), because the exponent in $e^{j2\pi \tau(\nu-\nu')}$ is positive. How should I modify these properties?

 

[blue_leaf77]

It seems like you have insufficient experience with Dirac delta functions.

For instance using (i), I think the second term in the equation becomes:

$$- \int^{\infty}_{-\infty} H(\nu') e^{2\pi j \nu' t} . \delta(\nu + \nu') d \nu'$$

Yes, that's right and to go further, do you know the sifting property of a delta function $\int_{-\infty}^\infty f(x) \delta(x-a) dx = f(a)$?

 

[roam]

It seems like you have insufficient experience with Dirac delta functions.

Yes, that's right and to go further, do you know the sifting property of a delta function $\int_{-\infty}^\infty f(x) \delta(x-a) dx = f(a)$?

Thank you for this hint. Unfortunately this is my first course on linear systems so I have very little experience with the mathematics involved.

So if I applied the sifting property correctly, by substituting the dummy variable $\nu'$ we obtain:

$$g(t) = \frac{1}{2j} \Big[ H(\nu') e^{2\pi j\nu' t} - H(-\nu') e^{-2\pi j \nu' t}\Big].$$

Using that property this can be written as: $\frac{1}{2j} \Big[ H(\nu') e^{2\pi j\nu' t} - H^*(\nu') e^{-2\pi j \nu' t}\Big].$

We were told to solve the problem this equation is what we need to derive. But how does this equation translate to $|H(\nu)| \sin \left( 2 \pi \nu t + arg(H(\nu)) \right)$?

Any explanation would be appreciated.

 

[blue_leaf77]

Using that property this can be written as: $\frac{1}{2j} \Big[ H(\nu') e^{2\pi j\nu' t} - H^*(\nu') e^{-2\pi j \nu' t}\Big].$

We were told to solve the problem this equation is what we need to derive. But how does this equation translate to $|H(\nu)| \sin \left( 2 \pi \nu t + arg(H(\nu)) \right)$?

You are now in the easy part actually. Express both $H(\nu)$ and $H^*(\nu)$ in a polar representation.

 

[Charles Link]

You almost have the final result. You need one more step and you should have the complete solution... One question you might be asking is, what does it do, and also what is the purpose of this seemingly lengthy formalism? It turns out, g(t) is the integral solution of an (inhomogeneous) linear differential equation that works from the time domain with a driving function f(t). h(t) is a response function that is associated with the differential equation. h(t) is also the solution to the differential equation when f(t) is a delta function $\delta(t)$ , and h(t) is also the (time) derivative of the response of the system when f(t) is a unit step function at t=0. It is perhaps a lot of work just to generate the Fourier transform result that $g(\omega)=H(\omega)*f(\omega)$ . You might find the electrical engineer's approach much simpler where you work with the ac circuit theory differential equation and you let $f(t)=V_{in}(t)=V_{in}exp(i \omega t)$ and assume in the steady state that $V_{out}(t)=V_{out}exp(i \omega t)$ where $V_{out}$ $(i.e. V_{out}(\omega))$ is a complex constant. In a simplified formalism, $V_{out}(\omega)=H(\omega)V_{in}(\omega)$ from which you can write $V_{out}(t)=H(\omega)V_{in}exp(i \omega t)$ and then take the real part of both sides of the final equation. A university physics professor once told our class that $exp(i \omega t)$ is essentially a poor man's Fourier transform. (It's a good shortcut, but perhaps lacking in mathematical rigor). The EE can often write the form of $H(\omega)$ by inspection using complex impedances for the different circuit elements. (e.g. $Z_L=i \omega L$ and $Z_C=-i /(\omega C)$ ) One additional item is if you use $exp(- i \omega t)$ instead of $exp(+i \omega t)$, it will change the sign of the complex impedances. (This basically makes the phasor diagrams go clockwise instead of counterclockwise. In some engineering books, you may see this convention). Hopefully you find this additional explanation helpful. The complete integral solution of these differential equations in the time domain can be a somewhat laborious process, but it can be extremely useful in some cases. The solution of even a simple R-C circuit with a voltage source using this more complete formalism (for the voltage across the capacitor) can be somewhat lengthy, but it does get the same result as the EE's shortcut solution that $V_{out}(t)=(- i /(\omega C))/(R-i/(\omega C))*V_{in}(t)$ where $V_{in}(t)=V_{in}*exp(i \omega t)$.