Baseband Equivalent of Passband Signals

Deterministic Baseband & Passband Signals and their Bandwidths

In a typical communication system, the information bearing signal that we wish to transmit is a signal whose spectrum is zero outside of the frequency range of $[-B,B]$Hz. For example, an audio signal is a signal whose bandwidth is restricted to at most a few KHz. Such signals which primarily occupy a range of frequencies centered around 0Hz are called low-pass signals or baseband signals. We will define the bandwidth of such a baseband signal as the smallest value of $W$ such that the spectrum is zero outside of the range $[-W,W]$Hz. It is important to pay attention to this definition of bandwidth. An example spectrum of such a baseband signal is shown in the figure below. Notice that according to our definition, the bandwidth is $B$ which is not the length of the support of $S(f)$.


For various reasons, information bearing baseband signals are usually up-converted using a carrier frequency of $f_c$ Hz and transmitted. Then, one obtains a signal $s(t)$ whose spectrum $S(f)$ is zero outside of the frequency range $f_c-B < |f| < f_c+B$, where $f_c >> B$. Such a signal is called a passband signal. In our definition, $S(f)$ may be zero for some frequencies in the range, $f_c-B < |f| < f_c+B$. $f_c$ is called the carrier frequency or reference frequency. Sometimes, the term center frequency is used - but this is a bit misleading since $f_c$ does not have to be midpoint of the range of frequencies over which $S(f)$ is non-zero for $f > 0$.

The passband signals we will encounter in this course are typically either transmitted waveforms, received waveforms, noise processes or interfering signals. In all these cases, the passband signal corresponds to a physically realizable voltage that can be measured at the output of a receive antenna. Hence, we will only consider in real passband signals in this course. As a result, the spectrum of any passband signal we consider, $S(f)$, is symmetric around 0 Hz. However, notice that this does not mean that the non-zero part of $S(f)$ for $f > 0$ is symmetric around $f_c$ Hz. The spectrum of one such a signal is shown in the figure below.


The bandwidth of such a passband signal with a reference frequency of $f_c$ will be defined as the smallest value of $W$ such that $S(f)$ is zero outside of the frequency range $f_c- \frac{W}{2} < |f| < f_c+\frac{W}{2}$.

Comments about our definition of bandwidth: It may seem at first that our definition of bandwidth of passband signals is inconsistent with that for baseband signals. This is not the case and a good way to remember this is to think about real signals. Then, for both real baseband and real passband signals, the bandwidth would be the length of range of positive frequencies around 0 or $f_c$ Hz which is non-zero. Please see page 104 from Lapidoth's book if you need more explanation. When the baseband signal is complex, we have to take the larger of the range of positive or negative frequencies over which the spectrum is non-zero (since the range would have to be of equal length around 0 or $f_c$ Hz). Again, the case of complex passband signals is not of interest.

A consequence of our definition of bandwidth is that multiplying a baseband signal by a carrier doubles its bandwidth. That is, if $p(t)$ is a baseband waveform with a bandwidth of $B$Hz, then the signal $q(t) = p(t) \ \cos(2 \pi f_c t)$ is a signal whose bandwidth is $2B$.

Complex Baseband Equivalent of Passband Signals

Transmitted signals and received signals in a communication system are real passband signals. However, it is often convenient to convert them to an equivalent baseband signal which preserves the information content in the passband signal and then process the baseband signal in order to recover the information (i.e., perform demodulation, equalization, etc). Given a real passband signal $s(t)$, there is an equivalent baseband signal $s_l(t)$ such that $s(t)$ can be obtained from $s_l(t)$ through some transformation (or up-conversion) process. Such a baseband signal $s_l(t)$ is called the baseband equivalent or low-pass equivalent of the passband signal $s(t)$.

To obtain such a signal, let us split $S(f)$ into the sum of two parts namely $S_+(f) = S(f) U(f)$ and $S_-(f) = S(f) U(-f)$ which represent the parts of $S(f)$ for positive frequencies and negative frequencies, respectively. We will define the baseband equivalent of $s(t)$ as the signal $s_l(t)$ for which

\begin{equation} S_l(f) = 2 S_+(f+f_c). \end{equation}

The scaling constant 2 in the above expression is somewhat arbitrary and we could have defined the baseband equivalent as the signal $s_l(t)$ for which $S_l(f) = k S_+(f+f_c)$. Some text books will use $k = \sqrt{2}$.

While $s(t)$ is a real signal, $s_l(t)$ is a complex signal in general and, hence, this is often referred to as a complex baseband equivalent signal. The complex baseband equivalent contains all the information in the real passband signal since the information is contained only in the shape of the $S(f)$ for positive frequencies. We will now explore the relationship between $s(t)$ and $s_l(t)$ in more detail.

$S(f)$ can be written in terms of $S_l(f)$ as

\begin{align} S(f)=\frac{1}{2}\left[S_l(f-f_c)+S_l^*(-f-f_c)\right] \end{align}

In the time domain, this implies that

\begin{eqnarray} s(t) &=& \frac{1}{2}\left[ s_l(t) e^{j 2 \pi f_c t}+s_l^*(t)e^{-j2\pi f_c t}\right] \\ &=& \Re \left[s_l(t) e^{j2\pi f_c t}\right] \end{eqnarray}

Thus, the signal $s_l(t)$ for which $s(t) = \Re \left[s_l(t) e^{j2\pi f_c t}\right]$ will be called the complex baseband equivalent of $s(t)$. While $s(t)$ is real, $s_l(t)$ is complex (in general).

Since $s_l(t)$ is complex, it can be written in Cartesian form as $s_l(t) = x(t) + j y(t)$. When this is used in 3, we can get the following expression for $s(t)$ in terms of $x(t)$ and $y(t)$.

\begin{align} s(t) = x(t) \cos(2 \pi f_c t) - y(t) \sin(2 \pi f_c t) \end{align}

The signals $x(t)$ and $y(t)$ are called the in-phase and quadrature components, respectively.

$s_l(t)$ can also be written in polar form as $s_l(t) = a(t) e^{j \theta(t)}$, where $a(t)$ is called the real envelope and $\theta(t)$ is called the excess phase. Then, $s(t)$ can be written in terms of $a(t)$ and $\theta(t)$ as

\begin{eqnarray} s(t) & = & \Re \{ a(t) e^{j(2 \pi f_c t+\theta(t))} \} \\ &=& a(t) \cos(2 \pi f_c t+\theta(t)) \end{eqnarray}

The above relationships provide a way to obtain $s(t)$ given an $s_l(t)$. Let us address the issue of finding the baseband equivalent $s_l(t)$ given a passband signal $s(t)$. Often, it is easy to do this by inspection or making a guess for $s_l(t)$ (or $x(t)$ and $y(t)$ such that $s(t) = \Re \left[s_l(t) e^{j2\pi f_c t}\right]$.

In an actual receiver the baseband equivalent is first obtained from the passband signal as follows. We first multiply $s(t)$ by $e^{-j 2 \pi f_c t}$ and then pass this signal through a low pass filter with a cut-off frequency of $B$Hz, where $B$ is the bandwith of the baseband signal. This produces the baseband equivalent. An implementation of the same idea is shown in the figure below where $x(t)$ and $y(t)$ are separately obtained.


We can also use the Hilbert transform to obtain $s_l(t)$ as described below. Let $S_+(f)=U(f)S(f)$ denote the part of $S(f)$ for $f > 0$. This is known as the analytic signal. Then, we can first obtain $s_+(t)$ from $s(t)$ according to

\begin{eqnarray} s_+(t) &=& F^{-1}[U(f)]*F^{-1}[S(f)] \\ &= & \frac{1}{2} \left[s(t)+\left( \frac{j}{\pi t} \right) *s(t) \right] \\ &=& \frac{1}{2} \left[s(t)+j \left( \frac{1}{\pi t} \right) *s(t) \right] \end{eqnarray}

and then we can move $S_+(f)$ to be centered around zero and multiply it by 2 to get $s_l(t)$ according to

\begin{align} s_l(t) = 2 s_+(t) e^{-j2\pi f_c t} \end{align}

A Hilbert transformer is simply a filter with impulse response $h(t)=\frac{1}{\pi t}$. Thus, we can obtain the imaginary part of $s_+(t)$ by passing $s(t)$ through such a Hilbert transformer. The name Hilbert transform is a bit of a misnomer as there is no transform of the signal to a new domain such as in a Fourier or Laplace transform. It is simply a filter with the given impulse response. The frequency response of the Hilbert transformer can be shown to be

\begin{eqnarray} H(f) &= &\int \frac{1}{\pi t} e^{-j2\pi f t}dt \\ &= & -j sgn(f) \\ & = & \left\{ \begin{array}{cc} -j & (f >0)\\ 0 & (f=0)\\j & (f<0) \end{array} \right. \end{eqnarray}

Let $\hat{s}(t)$ denote the Hilbert transform of $s(t)$. From the above relationship you can see that $\hat{S}(f) = -j sgn(f) S(f)$. Now, we can explicitly write $x(t)$ and $y(t)$ in terms of $s(t)$ and its Hilbert transform $\hat{s}(t)$. This can be done by first noting that

\begin{align} 2 s_+(t) = s(t) + j \hat{s}(t) \end{align}

Plugging this in to Eqn 10 and comparing the real and imaginary parts on both sides, we get

\begin{eqnarray} x(t) & = & s(t) \cos(2 \pi f_c t) + \hat{s}(t) \sin(2 \pi f_c t) \\ y(t) & = & \hat{s}(t) \cos (2 \pi f_c t) - s(t) \sin(2 \pi f_c t) \end{eqnarray}

A note on the complex baseband signal: Often students get confused by the fact that the baseband equivalent is complex since they try to interpret the baseband signal as some signal which they should be able to physically relate to. While passband signals are typically real physical signals, the equivalent complex baseband signal could be signal that exists only in our imagination.

If a passband signal is given, the baseband equivalent of the signal depends on the reference frequency $f_c$. The passband signal will result in different baseband signals $s_l(t)$ depending on the reference frequency $f_c$ we choose. Thus, it is always important to specify the reference frequency when talking about passband to baseband conversion.

Energy in Passband and Baseband

The energy in the passband signal $s(t)$ is given by,

\begin{eqnarray} E_S & = & \int_{-\infty}^{\infty}S^2(t)dt=\int \left[\frac{1}{2} S_l(t)e^{jw_ct}+\frac{1}{2} S_l^*(t) e^{-jw_ct} \right]^2dt \\ & = &\int\frac{1}{4} S_l^2(t)e^{j2w_ct}+\int\frac{1}{4} S_l^*^2(t)e^{-j2w_ct}+\frac{2}{4}\int|S_l(t)|^2dt\\ \\ &&\int S_l^2(t) e^{j2w_ct }dt = \int a(t) e^{j(2w_ct+2\theta (t)}dt \end{eqnarray}

We observe that the real part of the above integral is $\int a(t) \cos (2 \pi f_c t+2 \theta (t))$ which is zero since $a(t)$ is a signal that changes very slowly with time in relation to $\cos (2 \pi f_c t+2 \theta (t))$ and, hence, the integral is zero. Similarly, the second integral can also be seen to be zero. The key thing to realize: Both $a(t)$ and $\theta(t)$ are slowly varying processes compared to $cos(2 \omega _c t)$ and for this to be true, we require $f_c >> B/2$, the bandwidth of the low pass equivalent of the signal. This result is actually better seen in the frequency domain (need to add a frequency domain explanation for this). Thus,

\begin{align} E_s = \frac{1}{2} E_{s_l} = \frac{1}{2} \int_\infty ^\infty |S_l(t)|^2 dt \end{align}

The fact that $E_{s_l}$ is twice the energy of $E_s$ is a mere artifact of the way we have chosen to define the baseband equivalent in 1. If we had chosen to define the baseband equivalent as the signal $s_l(t)$ such that $s(t) = \sqrt{2} \re \{s_l(t) e^{j 2 \pi f_c t} \}$, then $E_s$ and $E_{s_l}$ would be the same.

Inner products in Passband and Baseband

Recall that for any two complex-valued (and trivially for real-valued) waveforms $s_{l1}(t)$ and $s_{l2}(t)$, the inner product between them is defined as

\begin{align} <s_{l1},s_{l2}> = \int s_{l1}(t) \ s_{l2}^* \ dt \end{align}

Suppose $s_1(t)$ and $s_2(t)$ are two passband signals with complex baseband equivalents $s_{l1}(t)$ and $s_{l2}(t)$, respectively, then the following relationship can be shown.

\begin{align} \int s_1(t) \ s_2^*(t) \ dt &=& \frac{1}{2} \Re \int s_{l1}(t) \ s_{l2}^*(t) \ dt \end{align}

Proof: Let us write the passband signals in terms of their baseband representations as

\begin{eqnarray} s_1(t)&=& \Re \{s_{l1}(t)e^{j2\pi f_c t}\} = \frac{1}{2} \left( s_{l1}(t)e^{j2\pi f_c t}+s_{l1}^*(t)e^{-j2 \pi f_c t} \right)\\ s_2(t)&=& \Re \{s_{l2}(t)e^{j2\pi f_c t}\} = \frac{1}{2} \left( s_{l2}(t)e^{j2\pi f_c t}+s_{l2}^*(t)e^{-j2 \pi f_c t} \right) \end{eqnarray}

Now, we can explicitly compute left hand side and using the fact that $\int s_{l1} s_{l2} e^{j 4 \pi f_c t} dt = \int s_{l1}^* s_{l2}^* e^{-j 4 \pi f_c t} dt = 0$, the desired result can be obtained. These results can be easily seen in the frequency domain by using Parseval's relation to compute these integrals in the frequency domain. It is a good idea to complete this proof at home and also think about what should be the relationship between $f_c$ and the bandwidth of $s_{l1}$, $s_{l2}$ for this to be true. Again, notice that the factor 1/2 in the relationship is a consequence of our definition of the baseband signal in .

An important consequence of this result is that

Orthogonality in baseband implies orthogonality in passband but the converse is not true.

Clearly, $s_1(t)$ and $s_2(t)$ are orthogonal, whereas $<s_{l1}(t),s_{l2}(t)> = -j$.

Baseband Equivalent Representation of Passband Filters

Just like how passband signals have an equivalent baseband representation, systems whose frequency response is passband in nature also have an equivalent baseband representation. This can be obtained by finding the baseband equivalent of the channel impulse response similar to the definition in 1.

\begin{eqnarray} H(f)&=&H_l(f-f_c)+H_l^*(-f-f_c)\\ h(t)&=& \Re \{h_l(t) e^{j\omega_c t}\}. \end{eqnarray}

The following figure shows the frequency of a passband system and the frequency response of its baseband equivalent.


Response of a Passband System to a Passband Signal

Often we are interested in understanding the effect of filters or linear time invariant systems on passband signals. Suppose the passband signal $s(t)$ is passed through an LTI system with an impulse response $h(t)$ with a frequency response which is also passband in nature and say the output is $r(t)$ whose baseband equivalent is $r_l(t)$. The relationship between $r(t)$, $s(t)$ and $h(t)$ is obvious. But what is the relationship between $r_l(t)$, $s_l(t)$ and $h_l(t)$. In this section, we will understand this.


We know,

\begin{eqnarray} R(f)&=&S(f) H(f) \\ &=& \frac{1}{2} \left[ S_l(f-f_c)+S_l^*(-f-f_c) \right] * \frac{1}{2} \left[ H_l(f-f_c)+ H_l^*(-f-f_c) \right] \\ &=&\frac{1}{4} \left[ S_l(f-f_c) H_l(f-f_c) + S_l^*(-f-f_c) H_l^*(-f-f_c) \right] +\frac{1}{4} \left[ S_l(f-f_c) H_l^*(-f-f_c) + S_l^*(-f-f_c) H_l(f-f_c) \right] \\ &=& \frac{1}{4} \left[ S_l(f-f_c) H_l(f-f_c) + S_l^*(-f-f_c) H_l^*(-f-f_c) \right] \end{eqnarray}

The last equation is obtained by noting the last two terms in the one before it are zero. If we define $R_l(f) = \frac{1}{2}S_l(f) H_l(f)$, then we can see that

\begin{align} R(f) = \frac{1}{2} \left[R_l(f-f_c) + R_l^*(-f-f_c)\right] \end{align}

which means that the signal $r_l(t) = \frac{1}{2} s_l(t) \star h_l(t)$ is indeed the baseband equivalent of $r(t)$.

Thus, we can just think about the filtering operation as being carried out in baseband except that the output of the baseband filter should be multiplied by 1/2 to obtain the baseband equivalent $r_l(t)$. This is demonstrated in the figure below.


Practice Problems

Here are a couple of problems that you can solve to improve your understanding of the relationship between bandpass signals and their lowpass equivalents.

  • Consider a real-valued signal $r(t)$ with Hilbert transform $\hat{r}(t)$. Let $r_I(t)$ and $r_Q(t)$ denote the in-phase and quadrature components of $r(t)$ with respect to a carrier frequency $f_o$. Let $x(t)$ and $y(t)$ be the in-phase and quadrature components of $\hat{r}(t)$ with respect to $f_o$. Express $x(t)$ and $y(t)$ in terms of $r_I(t)$ and $r_Q(t)$.
  • Let $s(t) = sin(1024 \pi t)/(1024 \pi t)$. Specify a particular carrier frequency $f_0$ in Hz such that the quadrature component of $s(t)$ with respect to $f_0$ is zero.
  • More exercises would be useful here.. Need to move homework problems to here.
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License