400*A*0.15=0.1 Watts

60*A=0.1 Watts

A=0.001667 meter squared

A=16.67 centimeter squared

Approximately 4cm x 4cm panel would do the job.

]]>▪ A radio of average power demand approximately 0.1 Watt.

For the bright light the power was 59.09 watts and the efficiency was (59.09/1)/400 = 0.15. The solar cell active area was 1m2.

How do I use this to solve the question?

Do you have the communication toolbox installed. Also, check the posts on OFDM, where I updated some code as I was also getting error messages.

Cheers ]]>

Thanks for your guide.

I tried to use the code you posted above but got this error

“Unable to resolve the name modem.qammod.”

Can you tell me the version of Matlab you are using?

1. So if you are using 250Watt panels you would need 20 of those (20 x 250Watts = 5000Watts)

2. Assuming Peak Sun Hours of 5 hours, energy produced would be 25KWhr (5hr x 5KW = 25KWhr)

3. If you want to use half of that energy and store half of that energy you will need 5 batteries of 200Ah each, at least

4. You will also need a medium sized inverter (depends upon load) and a charge controller

5. This system should be sufficient for a typical home with lights, fans, TV and refrigerator (but maybe not sufficient for an AC)

Hope if helps!

]]>Also, try different values of theta such as:

theta=0*pi/180

theta=45*pi/180

theta=90*pi/180

See how the polar plot changes. Basically you are changing the direction of the arriving wave.

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