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This lecture was delivered by Mr. Rohit Kohli at National Institute of Industrial Engineering for Embedded System Control course. It includes: Thalassemias, Blood, Hemoglobin, Oxygen, Heterozygous, Erythropoiesis, Hemoglobin, Homozygosity, Transfusions, Enormous
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Uploaded on 07/26/2012
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Simplex
Transmitter
Receiver
Half Duplex
Transmitter
Receiver
Receiver
Transmitter
Full Duplex
Transmitter
Receiver
Receiver
Transmitter
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EIA made interfacing standard in 1960.
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(Asynchronous)
Baud Rate = F
osc
/(64(X+1))
(Synchronous)
Baud Rate = F
osc
/(4(X+1))
BRGH = 1 (High Speed)
(Asynchronous)
Baud Rate = F
osc
/(16(X+1))
(Synchronous)
N/A
BRGH = 0 (Low Speed)
(Asynchronous)
=
F
osc
/(64 x baud rate)) -
(Synchronous)
=
F
osc
/(4 x baud rate)) -
BRGH = 1 (High Speed)
(Asynchronous)
=
F
osc
/(16 x baud rate)) -
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asynchronous mode high-speed transmission, assuming that the frequency of thecrystal oscillator is 20 MHz.
Solution:
Apply the formula The value (for BRGH = 1) to be written into the SPBRG
register is
SPBRG = (20,000,000 / (16 x 9600)) - 1 = 130 - 1 = 129 The actual baud rate is 20,000,000 / (16 x 130) = 9615.4 The resultant error rate is (9615.4 - 9600) / 9600 x 100% = 0.16% The same baud rate can also be achieved by using a low-speed (BRGH = 0) approach in
which
SPBRG = (20,000,000 / (64 + 9600)) - 1 = 31 The actual baud rate is 20000000 / (64 x 32) = 9765.6 The resultant error rate is (9765.6 - 9600) + 9600 x 100% = 1.7%
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SPEN
RX
SREN
CREN
ADDEN
FERR
OERR
RX9D
D
D
D
D
D
D
D
D
R
R
R
R
R
R
R
R docsity.com
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