Hilbert Transform Explained

The Hilbert transform of a signal $x\left(t\right)$ is

$$\boxed{\hat{x}(t)=\mathcal H\{x(t)\} =\frac{1}{\pi}\operatorname{PV}\int_{-\infty}^{\infty} \frac{x(\tau)}{t-\tau}\,d\tau}$$

where PV means the Cauchy principal value, because the kernel $\frac{1}{t-\tau }$ is singular at $\tau =t$.

Convolving a time-domain signal $f(t)$ with the Hilbert transform kernel $\frac{1}{\pi t}$ produces a 90° phase shift in its frequency-domain representation.

In the frequency domain:

$$\boxed{\mathcal F\{\hat{x}(t)\} =-j\,\operatorname{sgn}(\omega)X(\omega)}$$

So, intuitively, the Hilbert transform shifts each positive-frequency component by $-{90}^{\circ }$ and each negative-frequency component by $+{90}^{\circ }$, without changing its magnitude.

Why do we need it?

A very important application is creating the analytic signal:

$$z\left(t\right)=x\left(t\right)+j\mathcal{H}\left\{x\right(t\left)\right\}$$

For example, if

$$x\left(t\right)=\cos \left(\omega t\right)$$

then

$$z\left(t\right)=\cos \left(\omega t\right)+j\sin \left(\omega t\right)$$

and from Euler's formula,

$$z\left(t\right)=e^{j\omega t}$$

This is extremely useful for finding the instantaneous amplitude (envelope) and instantaneous phase of signals.

1. Start with Euler's formula

$$e^{j\theta }=\cos \theta +j\sin \theta$$

So we can write:

$$\cos \theta =\frac{e^{j\theta }+e^{-j\theta }}{2}$$

and

$$\sin \theta =\frac{e^{j\theta }-e^{-j\theta }}{2j}$$

The important point is that a real sine wave contains two frequencies:

• $+\omega$
• $-\omega$

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2. What does the Hilbert transform do?

In frequency domain, the Hilbert transform multiplies the spectrum by

$$-j\mathrm{sgn}\left(\omega \right)$$

That means:

For positive frequency $+\omega$:

$$-j$$

For negative frequency $-\omega$:

$$+j$$

And multiplying by $j$ means a $+{90}^{\circ }$ phase shift, while multiplying by $-j$ means a $-{90}^{\circ }$ phase shift.

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3. Apply it to sine

We have

$$\sin \left(\omega t\right)=\frac{e^{j\omega t}-e^{-j\omega t}}{2j}$$

Now apply the Hilbert transform.

For $e^{j\omega t}$, the frequency is positive, so multiply by $-j$:

$$e^{j\omega t}\to -j{e}^{j\omega t}$$

For $e^{-j\omega t}$, the frequency is negative, so multiply by $+j$:

$$e^{-j\omega t}\to +j{e}^{-j\omega t}$$

Therefore,

$$\mathcal{H}\left\{\sin \right(\omega t\left)\right\}=\frac{-j{e}^{j\omega t}+j{e}^{-j\omega t}}{2j}$$

Cancel $j$:

$$=\frac{-e^{j\omega t}+e^{-j\omega t}}{2}$$

Rearrange:

$$=-\frac{e^{j\omega t}+e^{-j\omega t}}{2}$$

But we know

$$\frac{e^{j\omega t}+e^{-j\omega t}}{2}=\cos \left(\omega t\right)$$

Therefore:

$$\boxed{\mathcal{H}\left\{\sin \right(\omega t\left)\right\}=-\cos \left(\omega t\right)}$$

The minus sign comes from treating positive and negative frequencies differently:

$$\boxed{+\omega :-j-\omega :+j}$$

That's the fundamental reason.

And similarly:

$$\boxed{\mathcal{H}\left\{\cos \right(\omega t\left)\right\}=\sin \left(\omega t\right)}$$

So don't memorize the two formulas separately. Remember the frequency-domain rule:

$$\boxed{X\left(\omega \right)\to -j\mathrm{sgn}\left(\omega \right)X\left(\omega \right)}$$

Everything else follows from that.

Conventional Carrier (Textbook Representation) vs. Real Passband Signal

A real cosine carrier is simply

$$\boxed{c\left(t\right)=\cos \left({\omega }_{c}t\right)}$$

This is a real physical waveform.

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Why does $j\sin$ appear?

We sometimes create a complex version of the carrier:

$$\boxed{e^{j{\omega }_{c}t}=\cos \left({\omega }_{c}t\right)+j\sin \left({\omega }_{c}t\right)}$$

This is Euler's formula.

There is nothing mysterious here.

It is just a convenient mathematical package containing:

$$\cos \left({\omega }_{c}t\right)$$

and its 90°-shifted version

$$\sin \left({\omega }_{c}t\right)$$

Think:

Complex carrier

       cosine
          ↓
e^(jωct) = cos(ωct) + j sin(ωct)
                         ↑
                       sine

The $j$ tells us that the sine component is the quadrature/90° component.

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Now suppose your information is $m\left(t\right)$

If you simply multiply it by the cosine carrier:

$$\boxed{s\left(t\right)=m\left(t\right)\cos \left({\omega }_{c}t\right)}$$

you get ordinary AM-type modulation.

No Hilbert transform is needed.

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What if we have I and Q?

Suppose we have two independent signals:

$$I\left(t\right)$$

and

$$Q\left(t\right)$$

Then we can form the complex baseband signal

$$\boxed{z\left(t\right)=I\left(t\right)+jQ\left(t\right)}$$

and use the complex carrier

$$e^{j{\omega }_{c}t}$$

So:

$$\boxed{z\left(t\right)e^{j{\omega }_{c}t}}$$

means

$$\boxed{\left[I\right(t)+jQ(t\left)\right]\left[\cos \right({\omega }_{c}t)+j\sin ({\omega }_{c}t\left)\right]}$$

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Where does the minus sign come from?

Expand it:

$$[I+jQ][\cos +j\sin ]$$

$$=I\cos +jI\sin +jQ\cos +j^{2}Q\sin$$

And because

$$\boxed{j^2=-1}$$

we get

$$=I\cos -Q\sin +j[I\sin +Q\cos ]$$

So:

$$\boxed{z\left(t\right)e^{j{\omega }_{c}t}=\left[I\cos \right({\omega }_{c}t)-Q\sin ({\omega }_{c}t\left)\right]+j\left[I\sin \right({\omega }_{c}t)+Q\cos ({\omega }_{c}t\left)\right]}$$

The real part is therefore

$$\boxed{s\left(t\right)=I\left(t\right)\cos \left({\omega }_{c}t\right)-Q\left(t\right)\sin \left({\omega }_{c}t\right)}$$

That's where the minus sign came from.

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Where does Hilbert transform enter?

Only now do we introduce it.

Suppose you have one real signal $m\left(t\right)$, but you want to construct a complex/analytic representation.

You calculate its Hilbert transform:

$$\boxed{\widehat{m}\left(t\right)=\mathcal{H}\left\{m\right(t\left)\right\}}$$

and form

$$\boxed{m_a\left(t\right)=m\left(t\right)+j\widehat{m}\left(t\right)}$$

So in this particular case:

$$\boxed{I\left(t\right)=m\left(t\right)}$$

and

$$\boxed{Q\left(t\right)=\widehat{m}\left(t\right)}$$

Then the real passband signal becomes

$$\boxed{s\left(t\right)=m\left(t\right)\cos \left({\omega }_{c}t\right)-\widehat{m}\left(t\right)\sin \left({\omega }_{c}t\right)}$$

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The whole picture

Keep this diagram:

                    REAL SIGNAL
                        m(t)
                         │
                         │ Hilbert transform
                         ▼
                       m̂(t)
                         │
                         │
              ┌──────────┴──────────┐
              │                     │
              ▼                     ▼
             I(t)                  Q(t)
             m(t)                 m̂(t)
              │                     │
              │ × cos              │ × sin
              ▼                     ▼
       I cos(ωct)             Q sin(ωct)
              │                     │
              └──────────┬──────────┘
                         │
                         ▼
             I cos(ωct) - Q sin(ωct)
                         │
                         ▼
                  REAL PASSBAND
                     SIGNAL

And separately:

Complex representation:

        z(t) = I(t) + jQ(t)

Carrier:

        e^(jωct)
        = cos(ωct) + j sin(ωct)

Together:

        z(t)e^(jωct)

The three equations you should remember

Complex carrier:

$$\boxed{e^{j{\omega }_{c}t}=\cos \left({\omega }_{c}t\right)+j\sin \left({\omega }_{c}t\right)}$$

Analytic signal:

$$\boxed{m_a\left(t\right)=m\left(t\right)+j\widehat{m}\left(t\right)}$$

Real passband signal:

$$\boxed{s\left(t\right)=m\left(t\right)\cos \left({\omega }_{c}t\right)-\widehat{m}\left(t\right)\sin \left({\omega }_{c}t\right)}$$