ECE538 Week 6: CTFT and D/A Conversion
The purpose of this week is to revisit the continuous time fourier transform (CTFT) to prepare to understand sampling theory and digital to analog (D/A) conversion.
CTFT
Definition
The CTFT of a signal $x(t)$ is defined as follows:
$$X(f) = \int_{t=-\infty}^{\infty}x(t)e^{-j2\pi ft}dt$$The inverse CTFT is given as:
$$x(t)=\int_{f=-\infty}^{\infty}X(f)e^{j2\pi ft}df$$Some quick notes:
- Notice that we are not using the $X(jw)$ notation
- Notice that we are in Hz, not rad/s. Things are much prettier this way :)
Side note:
This can be viewed as a projection of $x(t)$ onto $e^{j2\pi ft}$ in a Hilbert space with inner product $\langle x,y \rangle$ given as:
$$\langle x(t),y(t) \rangle=\int_{t=-\infty}^{\infty}x(t)y^*(t)dt$$Properties
Duality
From the definition of the CTFT, we can immediately obtain the following relation:
$$x(t) \overset{\mathcal{F}}{\leftrightarrow} X(f) \iff X(t) \overset{\mathcal{F}}{\leftrightarrow} x(-f)$$This is a useful property for obtaining many other properties.
Proof
Start with
$$X(f) = \int_{t=-\infty}^{\infty}x(t)e^{-j2\pi ft}dt \equiv x(t) \overset{\mathcal{F}}{\leftrightarrow} X(f)$$Rewrite the fourier transform with $f=z$, $t=y$
$$X(z) = \int_{y=-\infty}^{\infty}x(y)e^{-j2\pi yz}dy$$Now rewrite with $y = -f$ and $z = t$
$$X(t)=\int_{f=-\infty}^{\infty}x(-f)e^{j2\pi ft}df \equiv X(t) \overset{\mathcal{F}}{\leftrightarrow} x(-f)$$This means that
$$x(t) \overset{\mathcal{F}}{\leftrightarrow} X(f) \iff X(t) \overset{\mathcal{F}}{\leftrightarrow} x(-f)$$$$\fbox{QED}$$Linearity
It’s worth stating explicitly, but… duh…
$$a_1x_1(t)+a_2x_2(t) \overset{\mathcal{F}}{\leftrightarrow} a_1X_1(f)+a_2X_2(f)$$Time Scaling
This is a very useful property for conceptual understanding.
$$x(at) \overset{\mathcal{F}}{\leftrightarrow} \frac{1}{|a|}X(\frac{f}{a})$$As you can see, if you scale a signal horizontally in the time domain, the opposite occurs in the frequency domain.
This leads to a nice conceptual principle: $\fbox{The shorter the signal in time, the wider the signal in frequency.}$
And by duality, vice a versa.
Conjugation
$$x^*(t) \overset{\mathcal{F}}{\leftrightarrow} X^*(-f)$$Convolution/Multiplication
This is our first dual pair of properties. This means that once one result is achieved, the other can easily be through duality.
$$x_1(t)*x_2(t) \overset{\mathcal{F}}{\leftrightarrow} X_1(f)X_2(f)$$$$x_1(t)\cdot x_2(t) \overset{\mathcal{F}}{\leftrightarrow} X_1(f)*X_2(f)$$
Proof
$$\mathcal{F}\{x_1(t)*x_2(t)\}=\mathcal{F}\{\int_{\lambda=-\infty}^{\infty}x_1(\lambda)x_2(t-\lambda)d\lambda\}=\int_{\lambda=-\infty}^{\infty}x_1(\lambda)\mathcal{F}\{x_2(t-\lambda)\}d\lambda$$$$\mathcal{F}\{x_2(t-\lambda)\}=\int_{t=-\infty}^{\infty}x_2(t-\lambda)e^{-j2\pi ft}dt$$Let $u = t-\lambda$
$$\int_{t=-\infty}^{\infty}x_2(t-\lambda)e^{-j2\pi ft}dt = e^{-j2\pi f\lambda}\int_{u=-\infty}^{\infty}x_2(u)e^{-j2\pi fu}du = e^{-j2\pi f\lambda}X_2(f)$$Thus
$$\int_{\lambda=-\infty}^{\infty}x_1(\lambda)\mathcal{F}\{x_2(t-\lambda)\}d\lambda=X_2(f)\int_{\lambda=-\infty}^{\infty}x_1(\lambda)e^{-j2\pi f\lambda}d\lambda = X_1(f)X_2(f)$$Therefore
$$\mathcal{F}\{x_1(t)*x_2(t)\} = X_1(f)X_2(f)$$$$\fbox{QED pt 1}$$By duality:
$$X_1(t)\cdot X_2(t) \overset{\mathcal{F}}{\leftrightarrow} \int_{\lambda=-\infty}^{\infty} x_1(\lambda)x_2(-f-\lambda)d\lambda$$Let $\lambda = f - u$
$$\int_{\lambda=-\infty}^{\infty} x_1(\lambda)x_2(-f-\lambda)d\lambda = -\int_{u=\infty}^{-\infty} x_1(f - u)x_2(u)du$$(at this point, we could invoke commutativity of convolution and be done with it. However, we can push to the end)
Now let $u = f - v$
$$-\int_{u=\infty}^{-\infty} x_1(f - u)x_2(u)du = \int_{v=-\infty}^{\infty} x_1(v)x_2(f-v)dv = x_1(f)*x_2(f)$$Therefore
$$\mathcal{F}\{x_1(t)x_2(t)\} = X_1(f)*X_2(f)$$$$\fbox{QED pt 2}$$Complex Sinusoid / Delta
A fundamental pair:
$$\delta(t-t_0) \overset{\mathcal{F}}{\leftrightarrow} e^{-j2\pi ft_0}$$$$e^{j2\pi f_ct} \overset{\mathcal{F}}{\leftrightarrow} \delta(f-f_c)$$
Proof
Notice that we cannot attempt $\mathcal{F}\{e^{j2\pi f_ct}\}$ directly, as $e^{j2\pi f_ct}$ has infinite energy. However, we can take the other route and use duality to obtain the other.
$$\mathcal{F}\{\delta(t-t_0)\}=\int_{t=-\infty}^{\infty}\delta(t-t_0)e^{-j2\pi ft}dt=\int_{t=-\infty}^{\infty}\delta(t-t_0)e^{-j2\pi ft_0}dt=e^{-j2\pi ft_0}\int_{t=-\infty}^{\infty}\delta(t-t_0)dt=e^{-j2\pi ft_0}$$Therefore,
$$\mathcal{F}\{\delta(t-t_0)\}=e^{-j2\pi ft_0}$$$$\fbox{QED pt 1}$$Also notice
$$\mathcal{F}\{\delta(t+t_0)\}=e^{j2\pi ft_0}$$By duality, ($f_c = t_0$)
$$\mathcal{F}\{e^{j2\pi f_ct}\}=\delta(-f+f_c)=\delta(f-f_c)$$$$\fbox{QED pt 2}$$Time/Frequency Shifting
The below properties can be derived with a combination of the convolution property and duality.
$$x(t)e^{j2\pi f_ct} \overset{\mathcal{F}}{\leftrightarrow} X(f-f_c)$$$$x(t-t_0) \overset{\mathcal{F}}{\leftrightarrow} X(f)e^{-j2\pi ft_0}$$