1. Bandwidth part 1 What does it mean? Different things to different people.

2. The bandwidth of a signal is the range of frequencies it contains Bandwidth = f max – f min The bandwidth of this audio signal is about 3,500 Hz

3. Telephone call bandwidth = 4 kHz CD quality music bandwidth = 20 kHz Digital TV channel bandwidth = 8 MHz FM radio station bandwidth = 15 kHz

4. So I want to transmit Radio Drummy to the people What frequency radio waves do I transmit?

5. Frequency modulation I transmit at a very high frequency ( VHF ) e.g. 91 Mhz But modulate it slightly Change the frequency. The change in the frequency is my 15 kHz signal. I transmit 91 MHz + or – 7.5 kHz

6. Carrier frequency e.g. 100.8 MHz

8. The government owns the airwaves. You have to pay for bandwidth.

9. You also need a gap between stations. Why?

10. What bandwidth does a digital signal need? How could we send a digital signal on the airwaves?

11. We cant just send this An electromagnetic wave is made up of oscillations. The minimum frequency we need to send the above is this … Minimum frequency (bandwidth) needed = ½ bits / second

12. In practice we need higher frequency components As a rule of thumb … Minimum required bandwidth = Bit transmission rate

13. A digital TV channel is 8 million bits /sec. What bandwidth does it need? Could digital TV be transmitted in the VHF waveband?

14. Digital TV channels are transmitted at extremely high frequencies MICROWAVES e.g. between 30 and 31 GHz Why?

15. Bandwidth part 2

16. Bandwidth of a signal = the range of frequencies it contains, the width of its spectrum W

17. Minimum sampling rate needed (Nyquist rate) = 2 x highest frequency contained in signal = 2 W Bit rate = sampling rate x bits per sample = 2 W b

18. Bandwidth needed to transmit a digital signal (how much of the waveband we need) Minimum bandwidth needed = ½ bit rate In practice we need more than this Bandwidth needed = bit rate = B = 2Wb

19. The amount of noise in a signal limits the number of bits per sample b. Why?

20. Noise is greater than difference between sampling levels. Sampled bits include noise information. Noise falls between sampling levels To avoid sampling noise Maximum levels = 2 b = V total / V noise

21. e.g. if we are sampling a varying voltage of amplitude 200mV at 8 bits per sample how much noise is tolerable? V noise = V total / 256 = 0.781 mV