U-Net conv_block

Course Q&A Deep Learning Specialization Convolutional Neural Networks
1

Hello

I have the following question from U-Net cone-block.

The function for the conv_block is

conv_block(inputs=None, n_filters=32, dropout_prob=0, max_pooling=True)

The output is as below

input_size=(96, 128, 3)
n_filters = 32
inputs = Input(input_size)
cblock1 = conv_block(inputs, n_filters * 1)
model1 = tf.keras.Model(inputs=inputs, outputs=cblock1)

output1 = [[‘InputLayer’, [(None, 96, 128, 3)], 0],
[‘Conv2D’, (None, 96, 128, 32), 896, ‘same’, ‘relu’, ‘HeNormal’],
[‘Conv2D’, (None, 96, 128, 32), 9248, ‘same’, ‘relu’, ‘HeNormal’],
[‘MaxPooling2D’, (None, 48, 64, 32), 0, (2, 2)]]

My question is how is the filter size 896, 9248 computed in the output?

Thank you so much

Arunima

2

Hi Arunima,

896 and 9248 refer to the number of parameters calculated for the layer.

The first Conv2D is convolving 32 filters with size 3x3x3. This gives 32x3x3x3 + 32 (bias) parameters = 896 parameters.

The second Conv2D is convolving 32 filters with size 3x3x32 (to match the output volume of the previous Conv2D). This gives 32x3x3x32 + 32 parameters = 9248 parameters.

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3

Thank you so much for your response.

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