SME
Signals & Systems · Problem type
DFT bin spacing
Use this for “A 256-point DFT is taken of samples at 1 kHz. How far apart are the bins, and where does 100 Hz fall?”
Diagram
k = 25.6 on the top line sits over 100 Hz on the bottom line: each 1 above is 3.9063 below.
This is an example. Type your own values to start a new problem.
Δf (Hz)
=
fₛ (Hz)
÷
N
k
=
f (Hz)
÷
Δf (Hz)
T
Record length
calculated
Formulas
Δf
=
fₛ
N
3.9063
=
1,000
256
T
=
N
fₛ
0.256
=
256
1,000
k
=
f
Δf
25.6
=
100
3.9063
Formulas are worked in s. The step-by-step shows the conversions.
Assumptions
•
An N-point DFT splits 0 to fₛ into N bins, Δf = fₛ ÷ N apart; bin k sits at kΔf.
•
The record lasts T = N ÷ fₛ, so Δf = 1 ÷ T: a longer record gives finer bins.
•
A tone between bins (k not whole) leaks into the bins around it.
Step-by-step
WE KNOW
fₛ = 1,000 Hz, N = 256, f = 100 Hz
FIND
bin spacing (Δf), record length (T), bin position (k)
Step 1 · Find bin spacing (Δf)
Use Δf = fₛ ÷ N
The
N
bins
share
the
band
from
0
to
fₛ
equally.
Δf
=
1,000
256
Δf = 3.9063 Hz
Step 2 · Find record length (T)
Use T = N ÷ fₛ
N
samples,
one
every
1
fₛ
seconds,
make
the
record.
T
=
256
1,000
T
=
0.256
s
Step 3 · Find bin position (k)
Use k = f ÷ Δf
Count
the
bin
spacings
up
to
the
tone:
k
is
not
whole,
so
it
falls
between
bins
and
leaks.
k
=
100
3.9063
k
=
25.6
Step 4 · Convert the answers
Change each answer back to the unit you chose.
T = 0.256 s = 256 ms (1 s = 1000 ms)
Check
Put the numbers back in. Both sides should match.
3.9063
=
1,000
256
✓
0.256
=
256
1,000
✓
25.6
=
100
3.9063
✓
Topic check
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