Consider a linear time-variant channel. The transmitted signal is $x(t)$, the channel impulse response is $h(t, \tau)$, and the received signal is $y(t)$. Then $$ y(t) = \int_{-\infty}^\infty x(\tau) h(t, \tau) d\tau = x(t) \circledast h(t, \tau), \tag{1} $$ where $\circledast$ is the convolution operation. If the multipath channel is assumed to be a bandlimited bandpass channel, which is reasonable, then $h(t, \tau)$ may be equivalently described by a complex baseband impulse response $h_b(t, \tau)$ with the input and output being the complex envelope representations of the transmitted and received signals, respectively. That is, $$ r(t) = c(t) \circledast \frac{1}{2} h_b(t, \tau), \tag{2} $$ where $c(t)$ and $r(t)$ are the complex envelopes of $x(t)$ and $y(t)$, defined as $$ x(t) = \text{Re} \{ c(t) \exp(j 2 \pi f_c t) \} \tag{3} $$ and $$ y(t) = \text{Re} \{ r(t) \exp(j 2 \pi f_c t) \}. \tag{4} $$ My question is: How to derive (2)? Thanks in advance.

The following is my attempt to derive (2):

In terms of pre-envelopes, we have $x(t) = \text{Re} [ x_+(t) ]$ and $h(t, \tau) = \text{Re} [ h_+(t, \tau) ]$, where $$ x_+(t) = x(t) + j \hat{x}(t), \tag {5} $$ where $\hat{x}(t)$ is the Hilbert transform of signal $x(t)$, and $$ h_+(t, \tau) = h(t, \tau) + j \hat{h}(t, \tau) \tag {6}, $$ where $\hat{h}(t, \tau)$ is the Hilbert transform of $h(t, \tau)$. Then $$ y(t) = \int_{-\infty}^\infty \text{Re}[x_+(\tau)] \text{Re}[h_+(t, \tau)] d\tau. \tag{7} $$

In order to continue the derivation, we need the following lemma:

Lemma 1: $$ \int_{-\infty}^\infty \text{Re}[x_+(\tau)] \text{Re}[h_+(t, \tau)] d\tau = \frac{1}{2} \text{Re} \left[ \int_{-\infty}^\infty x_+(\tau) h_+^*(t, \tau) d\tau \right]. \tag{8} $$

Proof: We prove Lemma 1 by beginning with the right hand side (RHS) of (8):

\begin{align} \frac{1}{2} \text{Re} \left[ \int_{-\infty}^\infty x_+(\tau) h_+^*(t, \tau) d\tau \right] \notag &= \frac{1}{2} \text{Re} \left\{ \int_{-\infty}^\infty [ x(\tau) + j \hat{x}(\tau) ] [ h(t, \tau) - j \hat{h}(t, \tau) ] d\tau \right\} \notag\\ &= \frac{1}{2} \int_{-\infty}^\infty \text{Re} \left\{ [ x(\tau) + j \hat{x}(\tau) ] [ h(t, \tau) - j \hat{h}(t, \tau) ] \right\} d\tau \notag\\ &= \frac{1}{2} \int_{-\infty}^\infty [ x(\tau) h(t, \tau) + \hat{x}(\tau) \hat{h}(t, \tau) ] d\tau, \tag{9} \end{align} where $$ \int_{-\infty}^\infty \hat{x}(\tau) \hat{h}(t, \tau) d\tau = \int_{-\infty}^\infty \hat{x}(\tau) [\hat{h}^*(t, \tau)]^* d\tau. \tag{10} $$ Using the Plancherel theorem, (10) becomes $$ \int_{\infty}^\infty [ - j\ \text{sgn}(f) X(f) ] \{ [ - j\ \text{sgn}(-f) H(t, -f) ]^* \}^* df, \tag{11} $$ where $X(f)$ and $H(t, f)$ are Fourier Transforms of $x(t)$ and $h(t, \tau)$, respectively. (11) can be further simplified as \begin{align} \int_{-\infty}^\infty (-j)^2 \text{sgn}(f) \text{sgn}(-f) X(f) H(t, -f) df &= \int_{-\infty}^\infty X(f) H(t, -f) df \\ &= \int_{-\infty}^\infty X(f) [ H^*(t, -f) ]^* df. \tag{12} \end{align} Using Plancherel theorem again, we have \begin{align} (12) & = \int_{-\infty}^\infty x(t) [ h^*(t, \tau) ]^* d\tau \\ & = \int_{-\infty}^\infty x(t) h(t, \tau) d\tau. \tag{13} \end{align} Thus, \begin{align} (9) & = \frac{1}{2} \cdot 2 \int_{-\infty}^\infty x(t) h(t, \tau) d\tau \\ & = \int_{-\infty}^\infty x(t) h(t, \tau) d\tau \\ & = \int_{-\infty}^\infty \text{Re}[x_+(\tau)] \text{Re}[h_+(t, \tau)] d\tau. \tag{14} \end{align} This completes the proof of Lemma 1. $\square$

Using Lemma 1, we can rewrite $y(t)$ as follows: \begin{align} y(t) & = \frac{1}{2} \text{Re} \left[ \int_{-\infty}^\infty x_+(\tau) h_+^*(t, \tau) d\tau \right] \\ & = \frac{1}{2} \text{Re} \left[ \int_{-\infty}^\infty c(\tau) \exp(j 2 \pi f_c \tau) h_b^*(t, \tau) \exp(- j 2 \pi f_c t) d\tau \right] \\ & = \frac{1}{2} \text{Re} \left[ \exp(- j 2 \pi f_c t) \int_{-\infty}^\infty c(\tau) h_b^*(t, \tau) \exp(j 2 \pi f_c \tau) d\tau \right], \tag{15} \end{align} where the second equality in (15) results from (2.139) in [1]. Then I don't know how to continue...


[1] S. Haykin, Communication Systems, 3rd ed. John Wiley & Sons, Inc., 1994.

  • $\begingroup$ Your convolution notation rubs me the wrong way. If you have a convolution product, then the kernel only depends on one argument as it is translation invariant. Having two arguments indicates a more general integral transform that is not a convolution. Your integral and notation for $h$ should make that clear. But especially wrong seems to me to keep the arguments in the factors of what you say is a convolution product. The integration dummy $\tau$ should not appear anywhere outside the integral. $\endgroup$
    – Jazzmaniac
    Commented Oct 12, 2016 at 16:50

1 Answer 1


Note that in general, there is no guarantee that the output of a linear time-varying (LTV) system is a bandpass signal if the input is a bandpass signal. So for this discussion we need to restrict ourselves to systems for which we can assume that the output signal $y(t)$ is a bandpass signal if the input $x(t)$ also has that property. In this case we can use the complex notation for bandpass signals in a meaningful way:


where $x_a(t)$ is the analytic signal of $x(t)$, and $x_b(t)$ is the complex envelope of $x(t)$, with some appropriately chosen carrier frequency $\omega_c=2\pi f_c$. Similar definitions apply to the corresponding signals related to $y(t)$ and $h(t,\tau)$. Note the complex exponent as a function of $\tau$ (not $t$) in the complex notation of $h(t,\tau)$. If $h(t,\tau)$ described a linear time-invariant (LTI) system, it would be independent of $t$, and, consequently, it would only be a function of $\tau$ (which is the "memory" variable of the system).

I use $h(t,\tau)$ to denote the LTV system's input delay-spread function, which is sometimes also just referred to as "impulse response". It's important to be clear about its definition, because for LTV systems there are several equivalent time domain descriptions. With the definition I use here, $h(t,\tau)$ is the response at time $t$ to an impulse $\tau$ time units earlier. The input/output relation of the system is given by

$$y(t)=x(t)\star h(t,\tau)=\int_{-\infty}^{\infty}h(t,\tau)x(t-\tau)d\tau\tag{2}$$

This definition of $h(t,\tau)$ is usually used in the mobile/wireless communications literature. Note that it is slightly different from the notation you used in your question.

We define $x_a(t)$ by

$$x_a(t)=x(t)+j\mathcal{H}\{x(t)\}=x(t)+j(f\star x)(t)\tag{3}$$

where $\mathcal{H}\{\cdot\}$ denotes the Hilbert transform, and $f(t)=\text{p.v. }1/(\pi t)$ is the "impulse response" of an ideal Hilbert transformer. ["p.v." stand for "principal value", meaning that the Cauchy principal value is used to evaluate all implied integrals.] With $y_a(t)$ and $h_a(t,\tau)$ defined in a completely analogous manner, we can proceed by first proving the relation

$$y_a(t)=\frac12 x_a(t)\star h_a(t,\tau)\tag{4}$$

This can be done as follows (for the sake of readability I leave out the arguments $t$ and $\tau$):

$$\begin{align}\frac12(x_a\star h_a)&=\frac12(x+j(f\star x))\star(h+j(f\star h))\\&=\frac12[x\star h-\underbrace{(f\star f)}_{-\delta(t)}\star(x\star h)+2jf\star (x\star h)]\\&=x\star h+jf\star(x\star h)\\&=y+jf\star y\\&=y_a\end{align}$$

where I've used the fact that $\mathcal{H}\{\mathcal{H}\{x\}\}=-x$, i.e., $(f\star f)(t)=-\delta(t)$.

Using $(4)$ we can now easily prove the property

$$y_b(t)=\frac12 x_b(t)\star h_b(t,\tau)\tag{5}$$

With $y_a(t)=y_b(t)e^{j\omega_ct}$ (and similarly for $x_a(t)$ and $h_a(t,\tau)$), we get from $(4)$

$$\begin{align}y_b(t)e^{j\omega_ct}&=\frac12(x_b(t)e^{j\omega_ct})\star(h_b(t,\tau)e^{j\omega_c\tau})\\&=\frac12\int_{-\infty}^{\infty}h_b(t,\tau)e^{j\omega_c\tau}x_b(t-\tau)e^{j\omega_c(t-\tau)}d\tau\\&=\frac12\left[\int_{-\infty}^{\infty}h_b(t,\tau)x_b(t-\tau)d\tau\right]e^{j\omega_ct}\\&=\frac12 \left[x_b(t)\star h_b(t,\tau)\right]e^{j\omega_ct}\end{align}\tag{6}$$

from which $(5)$ follows immediately.

  • $\begingroup$ It is a simplified proof. You impose a constraint in your proof which is not given in the question. That is, you suppose the Hilbert transform exists. I can see by putting p.v. behind the definition of HT you want to extend the class of signals whose HT exist, but still the integral does not exist at the pole. $\endgroup$
    – msm
    Commented Oct 9, 2016 at 23:14
  • $\begingroup$ @PeterK. Even the OP uses frequency domain relations for the LTV system (eq. 11). He has provided two references in his question, if you are interested. $\endgroup$
    – msm
    Commented Oct 9, 2016 at 23:26
  • $\begingroup$ @msm: What we're trying to prove here is a relation between analytic signals, or, equivalently, between complex baseband representations of bandpass signals. These wouldn't even exist, if the corresponding Hilbert transforms didn't exist. So your objection is a bit like criticizing a proof of the Pythagorean theorem by saying "hey, what if there's no triangle?". $\endgroup$
    – Matt L.
    Commented Oct 10, 2016 at 7:16
  • $\begingroup$ @msm: Furthermore, the "p.v." is not there because I "want to extend the class of signals whose HT exist" but because it always needs to be there. It's not kind of trick I'm trying to use - as you seem to suggest -, but it's inherent in the definition of the Hilbert transform. Your claim "... but still the integral does not exist at the pole" is unintelligible. $\endgroup$
    – Matt L.
    Commented Oct 10, 2016 at 7:27
  • $\begingroup$ @msm For me to actually get your message, it needs to be made on the same answer as I commented on. @-ing someone who hasn't interacted with the answer you're commenting on doesn't work. $\endgroup$
    – Peter K.
    Commented Oct 10, 2016 at 14:36

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