The number of modes that can propagate in a rectangular waveguide is determined by solving for the cutoff frequency of each mode. The cutoff frequency for the $(m,n)$ mode in a rectangular waveguide is given by:

$f_{c(m,n)} = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}$

Where:

• $c$ is the speed of light in a vacuum ($3 \times 10^8$ m/s),
• $a$ is the wider dimension of the waveguide (2.5 cm = 0.025 m),
• $b$ is the narrower dimension of the waveguide (1 cm = 0.01 m),
• $m$ and $n$ are non-negative integers representing the mode numbers.

For any mode to propagate, the signal frequency $f$ (which is 20 GHz = $20 \times 10^9$ Hz) must be greater than the cutoff frequency $f_{c(m,n)}$.

Now, let's calculate the cutoff frequencies and find out which modes can propagate.The number of modes that can propagate in the waveguide for a 20 GHz signal is 6. Here are the first five modes with their respective cutoff frequencies:

1. $(0,1)$ mode: 15 GHz
2. $(1,0)$ mode: 6 GHz
3. $(1,1)$ mode: 16.16 GHz
4. $(2,0)$ mode: 12 GHz
5. $(2,1)$ mode: 19.21 GHz

All these modes have cutoff frequencies below 20 GHz, allowing them to propagate.

Would you like further details or have any questions? Here are some additional questions for further exploration:

1. What happens to modes with cutoff frequencies above the signal frequency?
2. How does changing the waveguide dimensions affect the number of modes?
3. What is the dominant mode in this waveguide?
4. How does the signal frequency influence waveguide performance?
5. Can higher-order modes cause signal interference in waveguides?

Tip: The TE$_{10}$ mode is the dominant mode in rectangular waveguides, as it usually has the lowest cutoff frequency.